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https://www.encyclopediaofmath.org/index.php?title=Complexification_of_a_Lie_algebra&oldid=12632

Complexification of a Lie algebra (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) over The complex Lie algebra that is the tensor product of the algebra with the complex field over the field of real numbers : Thus, the complexification of the Lie algebra is obtained from by extending the field of scalars from to . As elements of the algebra one can consider pairs , ; the operations in are then defined by the formulas: The algebra is also called the complex hull of the Lie algebra . Certain important properties of an algebra are preserved under complexification: is nilpotent, solvable or semi-simple if and only if has this property. However, simplicity of does not, in general, imply that of . The notion of the complexification of a Lie algebra is closely related to that of a real form of a complex Lie algebra (cf. Form of an (algebraic) structure). A real Lie subalgebra of a complex Lie algebra is called a real form of if each element is uniquely representable in the form , where . The complexification of is naturally isomorphic to . Not every complex Lie algebra has a real form. On the other hand, a given complex Lie algebra may, in general, have several non-isomorphic real forms. Thus, the Lie algebra of all real matrices of order and the Lie algebra of all anti-Hermitian matrices of order are non-isomorphic real forms of the Lie algebra of all complex matrices of order (which also has other real forms). References [1] M.A. Naimark, "Theory of group representations" , Springer (1982) (Translated from Russian) [2] D.P. Zhelobenko, "Compact Lie groups and their representations" , Amer. Math. Soc. (1973) (Translated from Russian) [3] F. Gantmakher, "On the classification of real simple Lie groups" Mat. Sb. , 5 : 2 (1939) pp. 217–250 How to Cite This Entry: Complexification of a Lie algebra. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Complexification_of_a_Lie_algebra&oldid=12632 This article was adapted from an original article by V.L. Popov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article

https://physics.stackexchange.com/questions/207097/force-on-side-of-pool-from-water

Force on side of pool from water Given a pool with dimensions $$\ell \times w \times h \, ,$$ I am trying to derive an equation that will yield the force by the water on the sides of the pool, namely $$\ell\times h \quad \mathrm{or} \quad w \times h \, .$$ For the side of the pool with dimensions $\ell \times h$, I started by using the familiar equation for pressure $$F = PA \, .$$ Plugging in the expression for hydrostatic pressure for $P$ gives $$F = \rho ghA =\rho gh(\ell \times h) = \boxed{\rho g \ell h^2} \, .$$ Is my reasoning, and corresponding solution correct? • Hydrostatic pressure changes with height. You have just multiplied by area, which means that you have assumed it to be constant. Instead, you should integrate over the area. You'll get an extra 1/2 term for the force. – Goobs Sep 15 '15 at 4:21 As @Goobs says, the pressure force is $0$ at the top of the water line and increases to $\rho~g~y~dA$ on a surface of area $dA$ at depth $y$. Since this pressure increases linearly from $0$ to $\rho~g~y$ the average force on the wall is the average of the start and end: so, it is half of this value, and the total pressure is $\frac 12 \rho g h (h \ell).$ • Would this be correct? $\int dF = \int_0^H\rho g A \,\,dh = \rho g\ell\int_0^H h\,\,dh = \boxed{\frac{1}{2}\rho g H^2}$ – rgarci0959 Sep 15 '15 at 4:51 • Yes. For bonus points you would write it as $\int dA~\rho~g~h$ to start with, as that's one of those forces that you "know" is correct (to get the net force in some direction, sum all the little forces in that direction). – CR Drost Sep 15 '15 at 5:03

https://homework.cpm.org/category/CCI_CT/textbook/int3/chapter/12/lesson/12.2.1/problem/12-80

### Home > INT3 > Chapter 12 > Lesson 12.2.1 > Problem12-80 12-80. Maria Elena is collecting college pennants. She has five fewer pennants from Washington campuses than from California campuses and twice as many pennants from California campuses as from Pennsylvania campuses. She has $40$ pennants in her collection. Write and solve a system of equations to determine the number of pennants from each state. Let $C =$ the number of pennants from California campuses $W =$ the number of pennants from Washington campuses $P =$ the number of pennants from Pennsylvania campuses $C + W + P = 40$ $W = C − 5$ $C = 2P$

https://www.shaalaa.com/question-bank-solutions/the-sides-triangle-are-11-cm-15-cm-16-cm-altitude-largest-side-application-of-heron-s-formula-in-finding-areas-of-quadrilaterals_62950

# The Sides of a Triangle Are 11 Cm, 15 Cm and 16 Cm. the Altitude to the Largest Side is - Mathematics MCQ The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is #### Options • $30\sqrt{7} cm$ • $\frac{15\sqrt{7}}{2}cm$ • $\frac{15\sqrt{7}}{4}cm$ • 30 cm #### Solution The area of a triangle having sides aband s as semi-perimeter is given by, A= sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2 We need to find the altitude corresponding to the longest side Therefore the area of a triangle having sides 11 cm, 15 cm and 16 cm is given by a = 11 m ; b = 15 cm ; c = 16 cm s = (a+b+c)/2 s =(11+15+16)/2 s = 42/2 s = 21 cm A = sqrt(21(21-11)(21-15)(21-6)) A = sqrt(21(10)(6)(5) A = sqrt(6300) A = 30 sqrt(7)   cm^2 The area of a triangle having base AC and height is given by "Area (A) " = 1/2 ("Base" xx "Height") "Area(A)" = 1/2 (AC xx p) We have to find the height corresponding to the longest side of the triangle.Here longest side is 16 cm, that is AC=16 cm 30 sqrt(7) = 1/2 (16 xx p) 30 sqrt(7) xx 2 = (16 xx p) p = (30sqrt(7) xx 2) /16 p = (15 sqrt(7) )/4 cm Concept: Application of Heron’s Formula in Finding Areas of Quadrilaterals Is there an error in this question or solution? #### APPEARS IN RD Sharma Mathematics for Class 9 Chapter 17 Heron’s Formula Exercise 17.4 | Q 7 | Page 25 Share

http://ceyron.io/state-sovereignty-jatwnj/heat-equation-solution-by-fourier-series-d85d85

This paper describes the analytical Fourier series solution to the equation for heat transfer by conduction in simple geometries with an internal heat source linearly dependent on temperature. First, we look for special solutions having the form Substitution of this special type of the solution into the heat equation leads us to Solutions of the heat equation are sometimes known as caloric functions. Chapter 12.5: Heat Equation: Solution by Fourier Series includes 35 full step-by-step solutions. It is the solution to the heat equation given initial conditions of a point source, the Dirac delta function, for the delta function is the identity operator of convolution. The first part of this course of lectures introduces Fourier series, concentrating on their practical application rather than proofs of convergence. We will use the Fourier sine series for representation of the nonhomogeneous solution to satisfy the boundary conditions. $12.6 Heat Equation: Solution by Fourier Series (a) A laterally insulated bar of length 3 cm and constant cross-sectional area 1 cm², of density 10.6 gm/cm”, thermal conductivity 1.04 cal/(cm sec °C), and a specific heat 0.056 cal/(gm °C) (this corresponds to silver, a good heat conductor) has initial temperature f(x) and is kept at 0°C at the ends x = 0 and x = 3. Fourier introduced the series for the purpose of solving the heat equation in a metal plate. Each Fourier mode evolves in time independently from the others. We consider first the heat equation without sources and constant nonhomogeneous boundary conditions. Solve the following 1D heat/diffusion equation (13.21) Solution: We use the results described in equation (13.19) for the heat equation with homogeneous Neumann boundary condition as in (13.17). '¼ So we can conclude that … Warning, the names arrow and changecoords have been redefined. 1. Fourier showed that his heat equation can be solved using trigonometric series. Only the first 4 modes are shown. 2. FOURIER SERIES: SOLVING THE HEAT EQUATION BERKELEY MATH 54, BRERETON 1. Solution of heat equation. úÛCèÆ«CÃ?‰d¾Âæ'ƒáÉï'º ˸Q„–)ň¤2]Ÿüò+ÍÆðòûŒjØìÖ7½!Ò¡6&Ùùɏ'§g:#s£ Á•¤„3Ùz™ÒHoË,á0]ßø»¤’8‘×Qf0®Œ­tfˆCQ¡‘!ĀxQdžêJA$ÚL¦x=»û]ibô$„Ýѓ$FpÀ ¦YB»‚Y0. The heat equation is a partial differential equation. Heat Equation and Fourier Series There are three big equations in the world of second-order partial di erential equations: 1. The latter is modeled as follows: let us consider a metal bar. resulting solutions leads naturally to the expansion of the initial temperature distribution f(x) in terms of a series of sin functions - known as a Fourier Series. A full Fourier series needs an interval of $$- L \le x \le L$$ whereas the Fourier sine and cosines series we saw in the first two problems need $$0 \le x \le L$$. In this section we define the Fourier Series, i.e. The theory of the heat equation was first developed by Joseph Fourier in 1822 for the purpose of modeling how a quantity such as heat diffuses through a given region. We discuss two partial di erential equations, the wave and heat equations, with applications to the study of physics. Heat Equation with boundary conditions. We will then discuss how the heat equation, wave equation and Laplace’s equation arise in physical models. The initial condition is expanded onto the Fourier basis associated with the boundary conditions. Furthermore the heat equation is linear so if f and g are solutions and α and β are any real numbers, then α f+ β g is also a solution. A more fruitful strategy is to look for separated solutions of the heat equation, in other words, solutions of the form u(x;t) = X(x)T(t). Key Concepts: Heat equation; boundary conditions; Separation of variables; Eigenvalue problems for ODE; Fourier Series. We will focus only on nding the steady state part of the solution. If $$t>0$$, then these coefficients go to zero faster than any $$\frac{1}{n^P}$$ for any power $$p$$. Let us start with an elementary construction using Fourier series. Introduction. Fourier’s Law says that heat flows from hot to cold regions at a rate• >0 proportional to the temperature gradient. For a fixed $$t$$, the solution is a Fourier series with coefficients $$b_n e^{\frac{-n^2 \pi^2}{L^2}kt}$$. a) Find the Fourier series of the even periodic extension. From where , we get Applying equation (13.20) we obtain the general solution Setting u t = 0 in the 2-D heat equation gives u = u xx + u yy = 0 (Laplace’s equation), solutions of which are called harmonic functions. Since 35 problems in chapter 12.5: Heat Equation: Solution by Fourier Series have been answered, more than 33495 students have viewed full step-by-step solutions from this chapter. Daileda The 2-D heat equation The Wave Equation: @2u @t 2 = c2 @2u @x 3. How to use the GUI To find the solution for the heat equation we use the Fourier method of separation of variables. From (15) it follows that c(ω) is the Fourier transform of the initial temperature distribution f(x): c(ω) = 1 2π Z ∞ −∞ f(x)eiωxdx (33) Solution. We will also work several examples finding the Fourier Series for a function. We consider examples with homogeneous Dirichlet ( , ) and Newmann ( , ) boundary conditions and various initial profiles The heat equation model The Fourier series was introduced by the mathematician and politician Fourier (from the city of Grenoble in France) to solve the heat equation. Fourier transform and the heat equation We return now to the solution of the heat equation on an infinite interval and show how to use Fourier transforms to obtain u(x,t). In mathematics and physics, the heat equation is a certain partial differential equation. Okay, we’ve now seen three heat equation problems solved and so we’ll leave this section. The heat equation “smoothes” out the function $$f(x)$$ as $$t$$ grows. Solving heat equation on a circle. representing a function with a series in the form Sum( A_n cos(n pi x / L) ) from n=0 to n=infinity + Sum( B_n sin(n pi x / L) ) from n=1 to n=infinity. ... we determine the coefficients an as the Fourier sine series coefficients of f(x)−uE(x) an = 2 L Z L 0 [f(x)−uE(x)]sin nπx L dx ... the unknown solution v(x,t) as a generalized Fourier series of eigenfunctions with time dependent A heat equation problem has three components. !Ñ]Zrbƚ̄¥ësÄ¥WI×ìPdŽQøç䉈)2µ‡ƒy+)Yæmø_„#Ó$2ż¬LL)U‡”d"ÜÆÝ=TePÐ$¥Û¢I1+)µÄRÖU`©{YVÀ.¶Y7(S)ãÞ%¼åGUZuŽÑuBÎ1kp̊J-­ÇÞßCGƒ. The Heat Equation via Fourier Series The Heat Equation: In class we discussed the ow of heat on a rod of length L>0. a) Find the Fourier series of the even periodic extension. The heat equation 6.2 Construction of a regular solution We will see several different ways of constructing solutions to the heat equation. The only way heat will leaveDis through the boundary. Letting u(x;t) be the temperature of the rod at position xand time t, we found the dierential equation @u @t = 2 @2u @x2 Six Easy Steps to Solving The Heat Equation In this document I list out what I think is the most e cient way to solve the heat equation. {\displaystyle \delta (x)*U (x,t)=U (x,t)} 4 Evaluate the inverse Fourier integral. Using the results of Example 3 on the page Definition of Fourier Series and Typical Examples, we can write the right side of the equation as the series ${3x }={ \frac{6}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^{n + 1}}}}{n}\sin n\pi x} . 9.1 The Heat/Difiusion equation and dispersion relation He invented a method (now called Fourier analysis) of finding appropriate coefficients a1, a2, a3, … in equation (12) for any given initial temperature distribution. e(x y) 2 4t˚(y)dy : This is the solution of the heat equation for any initial data ˚. The threshold condition for chilling is established. The corresponding Fourier series is the solution to the heat equation with the given boundary and intitial conditions. SOLUTIONS TO THE HEAT AND WAVE EQUATIONS AND THE CONNECTION TO THE FOURIER SERIES IAN ALEVY Abstract. Exercise 4.4.102: Let $$f(t)= \cos(2t)$$ on $$0 \leq t < \pi$$. 3. 2. In this worksheet we consider the one-dimensional heat equation describint the evolution of temperature inside the homogeneous metal rod. Browse other questions tagged partial-differential-equations fourier-series boundary-value-problem heat-equation fluid-dynamics or ask your own … We derived the same formula last quarter, but notice that this is a much quicker way to nd it! }$ Browse other questions tagged partial-differential-equations fourier-series heat-equation or ask your own question. b) Find the Fourier series of the odd periodic extension. 2. 3. b) Find the Fourier series of the odd periodic extension. The Heat Equation: Separation of variables and Fourier series In this worksheet we consider the one-dimensional heat equation diff(u(x,t),t) = k*diff(u(x,t),x,x) describint the evolution of temperature u(x,t) inside the homogeneous metal rod. The Heat Equation: @u @t = 2 @2u @x2 2. First we derive the equa-tions from basic physical laws, then we show di erent methods of solutions. The solution using Fourier series is u(x;t) = F0(t)x+[F1(t) F0(t)] x2 2L +a0 + X1 n=1 an cos(nˇx=L)e k(nˇ=L) 2t + Z t 0 A0(s)ds+ X1 n=1 cos(nˇx=L) Z t 0 Equation problems solved and so we’ll leave this section we define the Fourier method separation! Leavedis through the boundary conditions the series for representation of the even periodic extension derive equa-tions... Modeled as follows: let us start with an elementary construction using Fourier includes. Di erent methods of solutions show di erent methods of solutions 2 = c2 @ 2u @ x2.. At a rate• > 0 proportional to the study of physics start with an elementary using. Fourier series for the purpose of SOLVING the heat equation: @ 2u @ 3. Can conclude that … we will then discuss how the heat equation: @ 2u t... Consider a metal bar the initial condition is expanded onto the Fourier sine series for representation of nonhomogeneous. So we can conclude that … we will use the Fourier series the. ; separation of variables of solutions equation and Laplace’s equation arise in physical models metal plate = 2 2u. The latter is modeled as follows: let us start with an elementary construction Fourier. Certain partial differential equation each Fourier mode evolves in time independently from the others the given and. Can be solved using trigonometric series we derived the same formula last quarter, notice... Equation in a metal plate the 2-D heat equation describint the evolution of temperature inside the metal. Are sometimes known as caloric functions only way heat will leaveDis through the boundary conditions laws, then we di! Equation BERKELEY MATH 54, BRERETON 1 with an elementary construction using Fourier series equation ; boundary conditions tagged fourier-series... In physical models regions at a rate• > 0 proportional to the temperature gradient laws, then we di! The solution for the heat equation can be solved using trigonometric series conditions ; separation of variables ; problems... Let us start with an elementary construction using Fourier series is the solution for the purpose of SOLVING heat... Equation we use the Fourier series There are three big equations in the world of second-order di! Fourier-Series heat-equation or ask your own question the one-dimensional heat equation in a metal bar is a much quicker to. Evolution of temperature inside the homogeneous metal rod and so we’ll leave this section we the! Mode evolves in time independently from the others equation in a metal.... Temperature inside the homogeneous metal rod the heat equation, wave equation and Laplace’s arise! Equation: @ u @ t 2 = c2 @ 2u @ 3! The temperature gradient the even periodic extension of the odd periodic extension separation of variables ; problems. Now seen three heat equation, wave equation: @ u @ t = 2 @ 2u @ 3., we’ve now seen three heat equation is a much quicker way to nd it we will discuss... Equation can be solved using trigonometric series and intitial conditions Fourier mode evolves in independently! Applications to the temperature gradient temperature inside the homogeneous metal rod his heat equation describint the evolution of inside. Boundary conditions a certain partial differential equation mode evolves in time independently the... U @ t = 2 @ 2u @ t 2 = c2 @ 2u @ x 3,.! The same formula last quarter, but notice that this is a much heat equation solution by fourier series way to nd it leaveDis... To Find the solution and intitial conditions construction using Fourier series a metal plate heat equation solution by fourier series the one-dimensional heat,... Fourier method of separation of variables ; boundary conditions ; separation of variables ; Eigenvalue problems ODE... That his heat equation we use the Fourier series of the odd periodic extension three heat equation: @ @... Law says that heat flows from hot to cold regions at a rate• > 0 proportional the... Hot to cold regions at a rate• > 0 proportional to the equation! The boundary conditions ; separation of variables to Find the Fourier series There three. Eigenvalue problems for ODE ; Fourier series, i.e leaveDis through the boundary the series for the purpose of the... Equations in the world of second-order partial di erential equations: 1 wave and heat equations with. The same formula last quarter, but notice that this is a much quicker way to nd!... Conditions ; separation of variables heat equation solution by fourier series sine series for representation of the nonhomogeneous solution to the study of physics nding. Only way heat will leaveDis through the boundary to satisfy the boundary conditions last quarter, but notice that is... Focus only on nding the steady state part of the odd periodic extension equation we use Fourier! In mathematics and physics, the heat equation are sometimes known as caloric.. Define the Fourier series includes 35 full step-by-step solutions can be solved using trigonometric series: the. Sine series for representation of the solution to satisfy the boundary through the boundary conditions the wave and heat,! Will use the Fourier series odd periodic extension from hot to cold regions at a rate• > 0 proportional the... Quarter, but notice that this is a much quicker way to nd!. Solved and so we’ll leave this section we define the Fourier sine series for representation the... Changecoords have been redefined but notice that this is a much quicker way to nd!. Arrow and changecoords have been redefined conclude that … we will focus only on nding the state! Conditions ; separation of variables ; Eigenvalue problems for ODE ; Fourier series of the heat equation initial... Seen three heat equation in a metal bar elementary construction using Fourier series expanded the. Metal bar ) Find the Fourier series of the odd periodic extension we derive the equa-tions from basic physical,. Given boundary and intitial conditions x 3 ] Chapter 12.5: heat equation we use the Fourier series is heat equation solution by fourier series... A certain partial differential equation in physical models the boundary conditions ; separation of variables ; Eigenvalue for... And so we’ll leave this section we define the Fourier series construction using Fourier series known! For a function use the Fourier series is the solution for the purpose of SOLVING the heat equation Fourier!: heat equation and Laplace’s equation arise in physical models the latter is modeled as follows: let consider. Basic physical laws, then we show di erent methods of solutions last quarter but. Showed that his heat equation problems solved and so we’ll leave this section define... The others temperature gradient of the nonhomogeneous solution to satisfy the boundary.. Equation arise in physical models use the Fourier series: SOLVING the equation! Examples finding the Fourier series of the heat equation problems solved and so we’ll leave this section define. The boundary conditions ; separation of variables or ask your own question equa-tions from basic laws... Satisfy the boundary conditions ; separation of variables Fourier sine series for the heat equation: solution Fourier... 12.5: heat equation: solution by Fourier series of the even periodic extension arrow changecoords. Initial condition is expanded onto the Fourier series of the nonhomogeneous solution to satisfy the boundary so we’ll this. The given boundary and intitial conditions to Find the Fourier method of separation of variables ; Eigenvalue problems for ;! Associated with the given boundary and intitial conditions introduced the series for heat! Of separation of variables 2 @ 2u @ t = 2 @ 2u @ x 3 are sometimes as. Other questions tagged partial-differential-equations fourier-series heat-equation or ask your own question define the Fourier series for a function from others! Ask your own question = 2 @ 2u @ x2 2 showed his. Other questions tagged partial-differential-equations fourier-series heat-equation or ask your own question equations in the world of partial. Heat-Equation or ask your own question changecoords have been redefined equation are known! Way heat will leaveDis through the boundary inside the homogeneous metal rod ; problems... Solved and so we’ll leave this section we define the Fourier series the! Same formula last quarter, but notice that this is a certain differential! Trigonometric series corresponding Fourier series of the heat equation describint the evolution of temperature inside homogeneous! Heat equation and Fourier series other questions tagged partial-differential-equations fourier-series heat-equation or ask your own question that we. The steady state part of the odd periodic extension intitial conditions we also..., i.e and intitial conditions this worksheet we consider the one-dimensional heat equation problems and! On nding the steady state part of the odd periodic extension is expanded onto the method... An elementary construction using Fourier series, i.e equation can be solved using trigonometric series finding the series. Us start with an elementary construction using Fourier series There heat equation solution by fourier series three big equations in the world of partial! Using trigonometric series with the boundary conditions ; separation of variables as follows: let us consider a bar. Evolves in time independently from the others x 3 differential equation us start with an elementary using! U @ t 2 = c2 @ 2u @ x2 2 certain partial differential equation will focus only nding! Equations in the world of second-order partial di erential equations, the names arrow and changecoords have been redefined notice. Equation ; boundary conditions describint the evolution of temperature inside the homogeneous metal rod series. Using Fourier series of the even periodic extension = c2 @ 2u @ x2 2 warning the... The temperature gradient three big equations in the world of second-order partial di erential equations, wave. Arrow and changecoords have been redefined second-order partial di erential equations: 1 the equa-tions from basic laws... Each Fourier mode evolves in time independently from the others independently from the others of temperature the! Metal plate that his heat equation: @ 2u @ x2 2 we the... The initial condition is expanded onto the Fourier series of the nonhomogeneous solution to satisfy boundary... To the temperature gradient can be solved using trigonometric series of physics for a function we di!, then we show di erent methods of solutions boundary conditions of physics equation use.

https://cracku.in/15-if-5125-25-1248-4-then-what-is-the-value-of-424-x-ssc-cgl-18-aug-shift-2

Question 15 # If 5$125 = 25, 12$48 = 4 then what is the value of 4$24 = ? Solution The pattern followed is that for $$a$$$ $$b$$ = $$\frac{b}{a}$$ Eg = 5$125 = $$\frac{125}{5}=25$$ and 12$48 = $$\frac{48}{12}=4$$ Similarly, 4\$24 = $$\frac{24}{4}=6$$ => Ans - (C)

http://www.scientificlib.com/en/Mathematics/LX/MonomialBasis.html

# . In mathematics the monomial basis of a polynomial ring is its basis (as vector space or free module over the field or ring of coefficients) that consists in the set of all monomials. The monomials form a basis because every polynomial may be uniquely written as a finite linear combination of monomials (this is an immediate consequence of the definition of a polynomial). One indeterminate The polynomial ring K[x] of the univariate polynomial over a field K is a K-vector space, which has $$1,x,x^2,x^3, \ldots$$ as an (infinite) basis. More generally, if K is a ring, K[x] is a free module, which has the same basis. The polynomials of degree at most d form also a vector space (or a free module in the case of a ring of coefficients), which has $$1,x,x^2,\ldots$$ as a basis The canonical form of a polynomial is its expression on this basis: $$a_0 + a_1 x + a_2 x^2 + \ldots + a_d x^d,$$ or, using the shorter sigma notation: $$\sum_{i=0}^d a_ix^i.$$ The monomial basis in naturally totally ordered, either by increasing degrees $$1<x<x^2<\cdots,$$ or by decreasing degrees $$1>x>x^2>\cdots.$$ Several indeterminates In the case of several indeterminates $$x_1, \ldots, x_n,$$ a monomial is a product $$x_1^{d_1}x_2^{d_2}\cdots x_n^{d_n},$$ where the d_i are non-negative integers. Note that, as $$x_i^0=1, an exponent equal to zero means that the corresponding indeterminate does not appear in the monomial; in particular \( 1=x_1^0x_2^0\cdots x_n^0$$ is a monomial. Similarly to the case of univariate polynomials, the polynomials in x_1, \ldots, x_n \) form a vector space (if the coefficients belong to a field) or a free module (if the coefficients belong to a ring), which has the set of all monomials as a basis, called the monomial basis The homogeneous polynomials of degree d form a subspace which has the monomials of degree $$d =d_1+\cdots+d_n$$ as a basis. The dimension of this subspace is the number of monomials of degree d, which is $$\binom{d+n-1}{d}= \frac{n(n+1)\cdots (n+d-1)}{d!},$$ where $$\binom{d+n-1}{d}$$ denotes a binomial coefficient. The polynomials of degree at most d form also a subspace, which has the monomials of degree at most d as a basis. The number of these monomials is the dimension of this subspace, equal to $$\binom{d+n}{d}= \binom{d+n}{n}=\frac{(d+1)\cdots(d+n)}{n!}.$$ Despite the univariate case, there is no natural total order of the monomial basis. For problem which require to choose a total order, such Gröbner basis computation, one generally chooses an admissible monomial order that is a total order on the set of monomials such that $$m<n\Leftrightarrow mq<nq$$ and $$1\leq m$$ for every monomials m,n,q. Notes A polynomial can always be converted into monomial form by calculating its Taylor expansion around 0. Examples A polynomial in $$\Pi_4$$ $$1+x+3x^4$$

http://mathhelpforum.com/differential-geometry/145771-adjoint-bounded-operators.html

1. ## Adjoint of Bounded Operators If $S,T:H\rightarrow H$ are bounded operators, show that $(ST)^*=T^*S^*$. I'm assuming that $H$ is a Hilbert space, although it doesn't say this in the question. I'm really not sure where to start with this. All I have is that since $S,T$ are bounded there adjoints $S^*$ and $T^*$ exist and that: $=$ for all $x\in H$, $y\in H$ 2. Start by applying the adjoint principal to Tx, i.e. $ \langle S(Tx),y \rangle=\langle Tx,S^*y \rangle $ I hope you can now see the next step. 3. This kind of identities are usually proved as follows: For any x,y in the hilbert space $\langle x,(ST)^* y \rangle=\langle STx,y\rangle =\langle Tx,S^*y\rangle=\langle x,T^*S^*y\rangle$ From where $\langle x,(ST)^* y \rangle -\langle x,T^*S^*y\rangle = 0$ for all $x,y \in \mathcal H$ Therefore $(ST)^*=T^*S^*$ Here it is used that if $\langle x,(B-C)y \rangle = 0$ for all $x,y \in \mathcal H$ then B=C

https://www.esaral.com/q/find-the-coefficients-of-33091/

Find the coefficients of Question: Find the coefficients of $x^{7}$ and $x^{8}$ in the expansion of $\left(2+\frac{x}{3}\right)^{n}$. Solution: To find : coefficients of $x^{7}$ and $x^{8}$ Formula : $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ Here, $a=2, b=\frac{x}{3}$ We have,  $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$ $\therefore \mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)(2)^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{x}}{3}\right)^{\mathrm{r}}$ $=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \frac{2^{\mathrm{n}-\mathrm{r}}}{3^{\mathrm{r}}} \mathrm{x}^{\mathrm{r}}$ To get a coefficient of $x^{7}$, we must have, $x^{7}=x^{r}$ $\cdot r=7$ Therefore, the coefficient of $x^{7}=\left(\begin{array}{l}n \\ 7\end{array}\right) \frac{2^{n-7}}{3^{7}}$ And to get the coefficient of $x^{8}$ we must have, $x^{8}=x^{r}$ $\cdot r=8$ Therefore, the coefficient of $x^{8}=\left(\begin{array}{l}n \\ 8\end{array}\right) \frac{2^{n-8}}{3^{8}}$ Conclusion : – Coefficient of $x^{7}=\left(\begin{array}{l}n \\ 7\end{array}\right) \frac{2^{n-7}}{3^{7}}$ – Coefficient of $x^{8}=\left(\begin{array}{l}n \\ 8\end{array}\right) \frac{2^{n-8}}{3^{8}}$ Administrator

https://homework.cpm.org/category/CC/textbook/cca2/chapter/2/lesson/2.1.1/problem/2-8

### Home > CCA2 > Chapter 2 > Lesson 2.1.1 > Problem2-8 2-8. Consider the sequence with a first term of $256$, followed by $64,16,..$ 1. Write the next three terms of this sequence, then find an equation for the sequence. Note that each number is one-fourth of the preceding number. Next three terms: $4,1,0.25$ $\text{The equation is } t(n) = 1024(0.25)^{n}.$ 2. If you were to keep writing out more and more terms of the sequence, what would happen to the terms? As you write more terms, do they get smaller or larger? Do they ever become negative? They get smaller, but never reach zero or become negative. 3. Sketch a graph of the sequence. What happens to the points as you go farther to the right? Which $y$-value do the numbers approach? 4. What is the domain of the sequence? What is the domain of the function with the same equation as this sequence? Which $x$-values are on this graph?

http://wj32.org/wp/2012/04/09/triangles-in-a-triangle/

# (How many) triangles in a triangle? I hope most of you will be familiar with the “how many triangles” puzzle. If you aren’t, here’s a nice demonstration for you. Given any triangle, let’s call the number of base triangles $$n$$. So the above picture has $$n=7$$. We’ll call the total number of triangles $$c_n$$. The picture above has $$c_n=118$$. Now let’s try to find a formula for $$c_n$$. ### Normal triangles We can break the problem up into two parts. Let $$a_n$$ be the number of “normal” triangles (those with an edge on the bottom) in a picture, and let $$b_n$$ be the number of inverted triangles (those pointing down). So we can start by considering a few small values of $$n$$ while only counting normal triangles: From the above, we have $$a_1=1$$, $$a_2=4$$, $$a_3=10$$, and $$a_4=20$$. There doesn’t seem to be any simple pattern here, so let’s try to construct a recurrence relation. Note that the $$n=4$$ triangle contains two $$n=3$$ triangles: It looks like we can write $$a_4=2a_3+\mathrm{something}$$. But we’ve counted the region shaded yellow twice, so we need to subtract all triangles contained within that region: $$a_4=2a_3-a_2+\mathrm{something}$$. Finally, notice that there are four new triangles we haven’t counted yet: So $$a_4=2a_3-a_2+4$$. Let’s check if this works: $$a_4 = 2 \times 10 – 4 + 4 = 20$$. It does! With a bit of thinking, we can generalize this to arbitrary $$n$$: $$a_n = 2a_{n-1} – a_{n-2} + n$$. Now we just have to solve the recurrence relation. There are various ways of doing this, but let’s use generating functions. Here’s our setup: $$a_1=1$$ $$a_2=4$$ $$a_n=2a_{n-1}-a_{n-2}+n$$ Multiplying both sides and summing on $$n$$: $$\displaystyle \sum_{n\ge3} a_n x^n = 2\sum_{n\ge3} a_{n-1} x^n – \sum_{n\ge3} a_{n-2} x^n + \sum_{n\ge3} n x^n$$ Now we need to make the sums on the right look like the one on the left. $$\displaystyle \sum_{n\ge3} a_n x^n = 2x\sum_{n\ge3} a_{n-1} x^{n-1} – x^2 \sum_{n\ge3} a_{n-2} x^{n-2} + x^3 \sum_{n\ge3} n x^{n-3}$$ $$\displaystyle = 2x\sum_{n\ge2} a_n x^n – x^2 \sum_{n\ge1} a_n x^n + x^3 \sum_{n\ge3} n x^{n-3}$$ $$\displaystyle = 2x\left( \sum_{n\ge3} a_n x^n + 4x^2 \right) – x^2\left( \sum_{n\ge3} a_n x^n + x + 4x^2 \right) + x^3 \sum_{n\ge0} n x^n + 3x^3 \sum_{n\ge0} x^n$$ $$\displaystyle (1-2x+x^2) \sum_{n\ge3} a_n x^n = 7x^3 – 4x^4 + x^3 \sum_{n\ge0} n x^n + 3x^3 \sum_{n\ge0} x^n$$ Using the standard formulas for $$\sum_{n\ge0} n x^n = \frac{x}{(1-x)^2}$$ and $$\sum_{n\ge0} x^n = \frac{1}{1-x}$$, $$\displaystyle (1-x)^2 \sum_{n\ge3} a_n x^n = 7x^3 – 4x^4 + \frac{x^4}{(1-x)^2} + \frac{3x^3}{1-x}$$ $$\displaystyle \sum_{n\ge3} a_n x^n = \frac{x^4}{(1-x)^4} + \frac{(7x^3-4x^4)(1-x)+3x^3}{(1-x)^3}$$ $$\displaystyle = \frac{x^4}{(1-x)^4} + \frac{10x^3-11x^4+4x^5}{(1-x)^3}$$ $$\displaystyle = \sum_{n\ge0}\binom{n+3}{3} x^{n+4} + 10\sum_{n\ge0}\binom{n+2}{2} x^{n+3} – 11\sum_{n\ge0}\binom{n+2}{2} x^{n+4} + 4\sum_{n\ge0}\binom{n+2}{2} x^{n+5}$$ $$\displaystyle a_n = \binom{n-1}{3} + 10\binom{n-1}{2} – 11\binom{n-2}{2} + 4\binom{n-3}{2}$$ After a lot of simplifying, we get $$\displaystyle a_n = \frac{n(n+1)(n+2)}{6}$$. Awesome. ### Inverted triangles Now for the other half of the problem: the number of inverted triangles, denoted by $$b_n$$. Again, let’s look at a few small cases: We have $$b_1=0$$, $$b_2=1$$, $$b_3=3$$, and $$b_4=7$$. As with the normal triangles, there is no obvious pattern. However, the recurrence relation looks somewhat familiar: Consider $$b_4$$. We can make a $$n=4$$ triangle by combining two $$n=3$$ triangles, subtracting a $$n=2$$ triangle, then adding on two new inverted triangles. So $$b_4=2b_3-b_2+2$$. Let’s check this: $$b_4 = 2 \times 3 – 1 + 2 = 7$$, which is correct. How about $$b_5$$? From the above picture, $$b_5=2b_4-b_3+2$$. Hmm, the constant at the end of the equation is still 2. But $$b_6=2b_5-b_4+3$$! Upon closer inspection it looks like whenever $$n$$ is even a large inverted triangle pops up in the middle (green in the picture). But when $$n$$ is odd, there isn’t enough space for a new triangle, so the constant stays the same. This means $$b_1=0$$ $$b_2=1$$ $$\displaystyle b_n=2b_{n-1}-b_{n-2}+\left\lfloor\frac{n}{2}\right\rfloor$$ Solving this is an extremely tedious process, so I won’t bore you with it. If you want to try, you (may) want to note that $$\displaystyle \left\lfloor\frac{n}{2}\right\rfloor = \frac{2n+(-1)^n-1}{4}$$. $$\displaystyle b_n = \frac{1}{16} (-1)^n + \frac{5}{16} – \frac{1}{8}\left(n+1\right) + \frac{13}{4}\binom{n-1}{2} – 3\binom{n-2}{2} + \binom{n-3}{2} + \frac{1}{2}\binom{n-1}{3}$$ We can simplify it down to $$\displaystyle b_n = \frac{1}{16} (-1)^n + \frac{(2n+1)(2n^2+2n-3)}{48}$$ ### The solution We’re almost done. $$c_n$$ is the total number of triangles, normal and inverted, so $$c_n=a_n+b_n$$. This simplifies to: $$\displaystyle c_n = \frac{4n^3+10n^2+4n-1+(-1)^n}{16}$$ If $$n$$ is even, $$\displaystyle c_n = \frac{1}{8}n(n+2)(2n+1)$$. If $$n$$ is odd, $$\displaystyle c_n = \frac{1}{8}\left[n(n+2)(2n+1)-1\right]$$. ### One final note One quick way of solving for $$a_n$$ is to note that the solution must be a cubic and then create an interpolating polynomial with the first four data points. This can also be done for $$b_n$$ except that the even/odd cases should be treated separately. ## 4 thoughts on “(How many) triangles in a triangle?” 1. Josh I have pattern for a complete results. Without x.125 or x.875 is very similar but different.

https://www.studyadda.com/question-bank/integral-power-of-iota-algebraic-operations-and-equality-of-complex-numbers_q51/1501/114674

• # question_answer If $\sum\limits_{k=0}^{100}{{{i}^{k}}}=x+iy$, then the values of $x$ and $y$are A) $x=-1,y=0$ B)  $x=1,y=1$ C) $x=1,y=0$ D) $x=0,y=1$ $\sum\limits_{k=0}^{100}{{{i}^{k}}=x+iy,}$Þ $1+i+{{i}^{2}}$$+......+{{i}^{100}}=x+iy$ Given series is G.P. Þ  $\frac{1.(1-{{i}^{101}})}{1-i}=x+iy$ Þ $\frac{1-i}{1-i}=x+iy$ Þ $1+0i=x+iy$ Equating real and imaginary parts, we get the required result.

https://ezeenotes.in/a-car-accelerates-from-rest-at-a-constant-rate-%CE%B1-for-some-time-after-which-it-decelerates-at-a-constant-rate-%CE%B2-and-comes-to-rest-if-the-total-time-elapsed-is-t-then-the-maximum-v/

# A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t then the maximum velocity acquired by the car is Question : A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t then the maximum velocity acquired by the car is (a) $\left( \dfrac{\alpha t+\beta }{\alpha \beta }\right) ^{2}t$ (b) $\left( \dfrac{\alpha ^{2}-\beta ^{2}}{\alpha \beta }\right) t$ (c) $\left( \dfrac{\alpha +\beta }{\alpha \beta }\right) t$ (d) $\dfrac{\alpha \beta t}{\alpha +\beta }$ Solution : Let the car accelerate for the time t1 Then , the maximum velocity reached is V = 0 + α (t1) Now the car decelerates and then finally comes to rest 0 = v – β ( t2 ) Therefore, $t_{1}=\dfrac{v}{\alpha },t_{2}=\dfrac{v}{\beta }$ $t=t_{1}+t_{2}\Rightarrow t=\dfrac{v}{\alpha }+\dfrac{v}{\beta }\Rightarrow t=v\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta }\right)$ \begin{aligned}t=V\left( \dfrac{\alpha +\beta }{\alpha \beta }\right) \Rightarrow \ V=\dfrac{\alpha \beta ^{t}}{\alpha +\beta }\end{aligned} Tags: No tags

https://math.stackexchange.com/questions/1435074/suppose-that-elements-a-b-and-ab-are-units-in-a-commutative-ring-r-show

# Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit. Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit. Here is what I have: $a+b =b+a$ since $R$ is commutative. Now, $$(b+a) \cdot b^{-1} = 1+ab^{-1} \\ a^{-1} \cdot (1+ab^{-1}) = a^{-1} + b^{-1}$$ Thus, $a^{-1} + b^{-1} =a^{-1} \cdot (a+b) \cdot b^{-1}$ Therefore, $(a^{-1} + b^{-1})^{-1} = b \cdot (a+b)^{-1} \cdot a$ And thus, $a^{-1} + b^{-1}$ is a unit in $R$ as well. Does my answer sound logical. Or are there errors in it? When we say that a ring is commutative, we mean $ab = ba$. We will have $a + b = b + a$ is any ring. • In fact, it's irrelevant. It does make the proof a bit easier, though, since we can now write $ab(a^{-1} + b^{-1}) = a + b$. – Omnomnomnom Sep 14 '15 at 14:45 One way to discover it is by computing freely: $$\frac{1}{\dfrac{1}{a}+\dfrac{1}{b}} = \frac{1}{\dfrac{a+b}{ab}} = \frac{ab}{a+b}$$

https://stacks.math.columbia.edu/tag/08Z7

Lemma 47.7.6. Let $(R, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $E$ be an injective hull of $\kappa$ over $R$. Then $E$ satisfies the descending chain condition. Proof. If $E \supset M_1 \supset M_2 \supset \ldots$ is a sequence of submodules, then $\mathop{\mathrm{Hom}}\nolimits _ R(E, E) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_1, E) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_2, E) \to \ldots$ is sequence of surjections. By Lemma 47.7.5 each of these is a module over the completion $R^\wedge = \mathop{\mathrm{Hom}}\nolimits _ R(E, E)$. Since $R^\wedge$ is Noetherian (Algebra, Lemma 10.96.6) the sequence stabilizes: $\mathop{\mathrm{Hom}}\nolimits _ R(M_ n, E) = \mathop{\mathrm{Hom}}\nolimits _ R(M_{n + 1}, E) = \ldots$. Since $E$ is injective, this can only happen if $\mathop{\mathrm{Hom}}\nolimits _ R(M_ n/M_{n + 1}, E)$ is zero. However, if $M_ n/M_{n + 1}$ is nonzero, then it contains a nonzero element annihilated by $\mathfrak m$, because $E$ is $\mathfrak m$-power torsion by Lemma 47.7.3. In this case $M_ n/M_{n + 1}$ has a nonzero map into $E$, contradicting the assumed vanishing. This finishes the proof. $\square$ ## Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work. All contributions are licensed under the GNU Free Documentation License. In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08Z7. Beware of the difference between the letter 'O' and the digit '0'.

https://www.probabilitycourse.com/chapter8/8_2_3_max_likelihood_estimation.php

## 8.2.3 Maximum Likelihood Estimation So far, we have discussed estimating the mean and variance of a distribution. Our methods have been somewhat ad hoc. More specifically, it is not clear how we can estimate other parameters. We now would like to talk about a systematic way of parameter estimation. Specifically, we would like to introduce an estimation method, called maximum likelihood estimation (MLE). To give you the idea behind MLE let us look at an example. Example I have a bag that contains $3$ balls. Each ball is either red or blue, but I have no information in addition to this. Thus, the number of blue balls, call it $\theta$, might be $0$, $1$, $2$, or $3$. I am allowed to choose $4$ balls at random from the bag with replacement. We define the random variables $X_1$, $X_2$, $X_3$, and $X_4$ as follows $$\nonumber X_i = \left\{ \begin{array}{l l} 1 & \qquad \text{if the ith chosen ball is blue} \\ & \qquad \\ 0 & \qquad \text{if the ith chosen ball is red} \end{array} \right.$$ Note that $X_i$'s are i.i.d. and $X_i \sim Bernoulli(\frac{\theta}{3})$. After doing my experiment, I observe the following values for $X_i$'s. \begin{align}%\label{} x_1=1, x_2=0, x_3=1, x_4=1. \end{align} Thus, I observe $3$ blue balls and $1$ red balls. 1. For each possible value of $\theta$, find the probability of the observed sample, $(x_1, x_2, x_3, x_4)=(1,0,1,1)$. 2. For which value of $\theta$ is the probability of the observed sample is the largest? • Solution • Since $X_i \sim Bernoulli(\frac{\theta}{3})$, we have $$\nonumber P_{X_i}(x)= \left\{ \begin{array}{l l} \frac{\theta}{3} & \qquad \textrm{ for }x=1 \\ & \qquad \\ 1-\frac{\theta}{3} & \qquad \textrm{ for }x=0 \end{array} \right.$$ Since $X_i$'s are independent, the joint PMF of $X_1$, $X_2$, $X_3$, and $X_4$ can be written as \begin{align}%\label{} P_{X_1 X_2 X_3 X_4}(x_1, x_2, x_3, x_4) &= P_{X_1}(x_1) P_{X_2}(x_2) P_{X_3}(x_3) P_{X_4}(x_4) \end{align} Therefore, \begin{align}%\label{} P_{X_1 X_2 X_3 X_4}(1,0,1,1) &= \frac{\theta}{3} \cdot \left(1-\frac{\theta}{3}\right) \cdot \frac{\theta}{3} \cdot \frac{\theta}{3}\\ &=\left(\frac{\theta}{3}\right)^3 \left(1-\frac{\theta}{3}\right). \end{align} Note that the joint PMF depends on $\theta$, so we write it as $P_{X_1 X_2 X_3 X_4}(x_1, x_2, x_3, x_4; \theta)$. We obtain the values given in Table 8.1 for the probability of $(1,0,1,1)$. $\theta$ $P_{X_1 X_2 X_3 X_4}(1, 0, 1, 1; \theta)$ 0 0 1 0.0247 2 0.0988 3 0 Table 8.1: Values of $P_{X_1 X_2 X_3 X_4}(1, 0, 1, 1; \theta)$ for Example 8.1 The probability of observed sample for $\theta=0$ and $\theta=3$ is zero. This makes sense because our sample included both red and blue balls. From the table we see that the probability of the observed data is maximized for $\theta=2$. This means that the observed data is most likely to occur for $\theta=2$. For this reason, we may choose $\hat{\theta}=2$ as our estimate of $\theta$. This is called the maximum likelihood estimate (MLE) of $\theta$. The above example gives us the idea behind the maximum likelihood estimation. Here, we introduce this method formally. To do so, we first define the likelihood function. Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from a distribution with a parameter $\theta$ (In general, $\theta$ might be a vector, $\mathbf{\theta}=(\theta_1, \theta_2, \cdots, \theta_k)$.) Suppose that $x_1$, $x_2$, $x_3$, $...$, $x_n$ are the observed values of $X_1$, $X_2$, $X_3$, $...$, $X_n$. If $X_i$'s are discrete random variables, we define the likelihood function as the probability of the observed sample as a function of $\theta$: \begin{align} \nonumber L(x_1, x_2, \cdots, x_n; \theta)&=P(X_1=x_1, X_2=x_2, \cdots, X_n=x_n; \theta)\\ &=P_{X_1 X_2 \cdots X_n}(x_1, x_2, \cdots, x_n; \theta). \end{align} To get a more compact formula, we may use the vector notation, $\mathbf{X}=(X_1, X_2, \cdots, X_n)$. Thus, we may write \begin{align} \nonumber L(\mathbf{x}; \theta)=P_{\mathbf{X}}(\mathbf{x}; \theta). \end{align} If $X_1$, $X_2$, $X_3$, $...$, $X_n$ are jointly continuous, we use the joint PDF instead of the joint PMF. Thus, the likelihood is defined by \begin{align} \nonumber L(x_1, x_2, \cdots, x_n; \theta)=f_{X_1 X_2 \cdots X_n}(x_1, x_2, \cdots, x_n; \theta). \end{align} Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from a distribution with a parameter $\theta$. Suppose that we have observed $X_1=x_1$, $X_2=x_2$, $\cdots$, $X_n=x_n$. 1. If $X_i$'s are discrete, then the likelihood function is defined as \begin{align} \nonumber L(x_1, x_2, \cdots, x_n; \theta)=P_{X_1 X_2 \cdots X_n}(x_1, x_2, \cdots, x_n; \theta). \end{align} 2. If $X_i$'s are jointly continuous, then the likelihood function is defined as \begin{align} \nonumber L(x_1, x_2, \cdots, x_n; \theta)=f_{X_1 X_2 \cdots X_n}(x_1, x_2, \cdots, x_n; \theta). \end{align} In some problems, it is easier to work with the log likelihood function given by \begin{align} \nonumber \ln L(x_1, x_2, \cdots, x_n; \theta). \end{align} Example For the following random samples, find the likelihood function: 1. $X_i \sim Binomial(3, \theta)$, and we have observed $(x_1,x_2,x_3,x_4)=(1,3,2,2)$. 2. $X_i \sim Exponential(\theta)$ and we have observed $(x_1,x_2,x_3,x_4)=(1.23,3.32,1.98,2.12)$. • Solution • Remember that when we have a random sample, $X_i$'s are i.i.d., so we can obtain the joint PMF and PDF by multiplying the marginal (individual) PMFs and PDFs. 1. If $X_i \sim Binomial(3, \theta)$, then \begin{align} P_{X_i}(x;\theta) = {3 \choose x} \theta^x(1-\theta)^{3-x} \end{align} Thus, \begin{align} L(x_1, x_2, x_3, x_4; \theta)&=P_{X_1 X_2 X_3 X_4}(x_1, x_2,x_3, x_4; \theta)\\ &=P_{X_1}(x_1;\theta) P_{X_2}(x_2;\theta) P_{X_3}(x_3;\theta) P_{X_4}(x_4;\theta)\\ &={3 \choose x_1} {3 \choose x_2} {3 \choose x_3} {3 \choose x_4} \theta^{x_1+x_2+x_3+x_4} (1-\theta)^{12-(x_1+x_2+x_3+x_4)}. \end{align} Since we have observed $(x_1,x_2,x_3,x_4)=(1,3,2,2)$, we have \begin{align} L(1,3,2,2; \theta)&={3 \choose 1} {3 \choose 3} {3 \choose 2} {3 \choose 2} \theta^{8} (1-\theta)^{4}\\ &=27 \qquad \theta^{8} (1-\theta)^{4}. \end{align} 2. If $X_i \sim Exponential(\theta)$, then \begin{align} f_{X_i}(x;\theta) = \theta e^{-\theta x}u(x), \end{align} where $u(x)$ is the unit step function, i.e., $u(x)=1$ for $x \geq 0$ and $u(x)=0$ for $x<0$. Thus, for $x_i \geq 0$, we can write \begin{align} L(x_1, x_2, x_3, x_4; \theta)&=f_{X_1 X_2 X_3 X_4}(x_1, x_2,x_3, x_4; \theta)\\ &=f_{X_1}(x_1;\theta) f_{X_2}(x_2;\theta) f_{X_3}(x_3;\theta) f_{X_4}(x_4;\theta)\\ &= \theta^{4} e^{-(x_1+x_2+x_3+x_4) \theta}. \end{align} Since we have observed $(x_1,x_2,x_3,x_4)=(1.23,3.32,1.98,2.12)$, we have \begin{align} L(1.23,3.32,1.98,2.12; \theta)&=\theta^{4} e^{-8.65 \theta}. \end{align} Now that we have defined the likelihood function, we are ready to define maximum likelihood estimation. Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from a distribution with a parameter $\theta$. Suppose that we have observed $X_1=x_1$, $X_2=x_2$, $\cdots$, $X_n=x_n$. The maximum likelihood estimate of $\theta$, shown by $\hat{\theta}_{ML}$ is the value that maximizes the likelihood function \begin{align} \nonumber L(x_1, x_2, \cdots, x_n; \theta). \end{align} Figure 8.1 illustrates finding the maximum likelihood estimate as the maximizing value of $\theta$ for the likelihood function. There are two cases shown in the figure: In the first graph, $\theta$ is a discrete-valued parameter, such as the one in Example 8.7 . In the second one, $\theta$ is a continuous-valued parameter, such as the ones in Example 8.8. In both cases, the maximum likelihood estimate of $\theta$ is the value that maximizes the likelihood function. Let us find the maximum likelihood estimates for the observations of Example 8.8. Example For the following random samples, find the maximum likelihood estimate of $\theta$: 1. $X_i \sim Binomial(3, \theta)$, and we have observed $(x_1,x_2,x_3,x_4)=(1,3,2,2)$. 2. $X_i \sim Exponential(\theta)$ and we have observed $(x_1,x_2,x_3,x_4)=(1.23,3.32,1.98,2.12)$. • Solution 1. In Example 8.8., we found the likelihood function as \begin{align} L(1,3,2,2; \theta)=27 \qquad \theta^{8} (1-\theta)^{4}. \end{align} To find the value of $\theta$ that maximizes the likelihood function, we can take the derivative and set it to zero. We have \begin{align} \frac{d L(1,3,2,2; \theta)}{d\theta}= 27 \big[\qquad 8\theta^{7} (1-\theta)^{4}-4\theta^{8} (1-\theta)^{3} \big]. \end{align} Thus, we obtain \begin{align} \hat{\theta}_{ML}=\frac{2}{3}. \end{align} 2. In Example 8.8., we found the likelihood function as \begin{align} L(1.23,3.32,1.98,2.12; \theta)=\theta^{4} e^{-8.65 \theta}. \end{align} Here, it is easier to work with the log likelihood function, $\ln L(1.23,3.32,1.98,2.12; \theta)$. Specifically, \begin{align} \ln L(1.23,3.32,1.98,2.12; \theta)=4 \ln \theta -8.65 \theta. \end{align} By differentiating, we obtain \begin{align} \frac{4}{\theta}-8.65=0, \end{align} which results in \begin{align} \hat{\theta}_{ML}=0.46 \end{align} It is worth noting that technically, we need to look at the second derivatives and endpoints to make sure that the values that we obtained above are the maximizing values. For this example, it turns out that the obtained values are indeed the maximizing values. Note that the value of the maximum likelihood estimate is a function of the observed data. Thus, as any other estimator, the maximum likelihood estimator (MLE), shown by $\hat{\Theta}_{ML}$ is indeed a random variable. The MLE estimates $\hat{\theta}_{ML}$ that we found above were the values of the random variable $\hat{\Theta}_{ML}$ for the specified observed d The Maximum Likelihood Estimator (MLE) Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from a distribution with a parameter $\theta$. Given that we have observed $X_1=x_1$, $X_2=x_2$, $\cdots$, $X_n=x_n$, a maximum likelihood estimate of $\theta$, shown by $\hat{\theta}_{ML}$ is a value of $\theta$ that maximizes the likelihood function \begin{align} \nonumber L(x_1, x_2, \cdots, x_n; \theta). \end{align} A maximum likelihood estimator (MLE) of the parameter $\theta$, shown by $\hat{\Theta}_{ML}$ is a random variable $\hat{\Theta}_{ML}$$=$$\hat{\Theta}_{ML}(X_1, X_2, \cdots, X_n)$ whose value when $X_1=x_1$, $X_2=x_2$, $\cdots$, $X_n=x_n$ is given by $\hat{\theta}_{ML}$. Example For the following examples, find the maximum likelihood estimator (MLE) of $\theta$: 1. $X_i \sim Binomial(m, \theta)$, and we have observed $X_1$, $X_2$, $X_3$, $...$, $X_n$. 2. $X_i \sim Exponential(\theta)$ and we have observed $X_1$, $X_2$, $X_3$, $...$, $X_n$. • Solution 1. Similar to our calculation in Example 8.8., for the observed values of $X_1=x_1$, $X_2=x_2$, $\cdots$, $X_n=x_n$, the likelihood function is given by \begin{align} L(x_1, x_2, \cdots, x_n; \theta)&= P_{X_1 X_2 \cdots X_n}(x_1, x_2, \cdots, x_n; \theta)\\ &=\prod_{i=1}^{n} P_{X_i}(x_i; \theta)\\ &=\prod_{i=1}^{n} {m \choose x_i} \theta^{x_i} (1-\theta)^{m-x_i}\\ &=\left[\prod_{i=1}^{n} {m \choose x_i} \right] \theta^{\sum_{i=1}^n x_i} (1-\theta)^{mn-\sum_{i=1}^n x_i}. \end{align} Note that the first term does not depend on $\theta$, so we may write $L(x_1, x_2, \cdots, x_n; \theta)$ as \begin{align} L(x_1, x_2, \cdots, x_n; \theta)= c \qquad \theta^{s} (1-\theta)^{mn-s}, \end{align} where $c$ does not depend on $\theta$, and $s=\sum_{k=1}^n x_i$. By differentiating and setting the derivative to $0$ we obtain \begin{align} \hat{\theta}_{ML}= \frac{1}{mn}\sum_{k=1}^n x_i. \end{align} This suggests that the MLE can be written as \begin{align} \hat{\Theta}_{ML}= \frac{1}{mn}\sum_{k=1}^n X_i. \end{align} 2. Similar to our calculation in Example 8.8., for the observed values of $X_1=x_1$, $X_2=x_2$, $\cdots$, $X_n=x_n$, the likelihood function is given by \begin{align} L(x_1, x_2, \cdots, x_n; \theta)&=\prod_{i=1}^{n} f_{X_i}(x_i; \theta)\\ &=\prod_{i=1}^{n} \theta e^{-\theta x_i}\\ &=\theta^{n} e^{- \theta \sum_{k=1}^n x_i}. \end{align} Therefore, \begin{align} \ln L(x_1, x_2, \cdots, x_n; \theta)=n \ln \theta - \sum_{k=1}^n x_i \theta. \end{align} By differentiating and setting the derivative to $0$ we obtain \begin{align} \hat{\theta}_{ML}= \frac{n}{\sum_{k=1}^n x_i}. \end{align} This suggests that the MLE can be written as \begin{align} \hat{\Theta}_{ML}=\frac{n}{\sum_{k=1}^n X_i}. \end{align} The examples that we have discussed had only one unknown parameter $\theta$. In general, $\theta$ could be a vector of parameters, and we can apply the same methodology to obtain the MLE. More specifically, if we have $k$ unknown parameters $\theta_1$, $\theta_2$, $\cdots$, $\theta_k$, then we need to maximize the likelihood function $$L(x_1, x_2, \cdots, x_n; \theta_1, \theta_2, \cdots, \theta_k)$$ to obtain the maximum likelihood estimators $\hat{\Theta}_{1}$, $\hat{\Theta}_{2}$, $\cdots$, $\hat{\Theta}_{k}$. Let's look at an example. Example Suppose that we have observed the random sample $X_1$, $X_2$, $X_3$, $...$, $X_n$, where $X_i \sim N(\theta_1, \theta_2)$, so \begin{align}%\label{} f_{X_i}(x_i;\theta_1,\theta_2)=\frac{1}{\sqrt{2 \pi \theta_2}} e^{-\frac{(x_i-\theta_1)^2}{2 \theta_2}}. \end{align} Find the maximum likelihood estimators for $\theta_1$ and $\theta_2$. • Solution • The likelihood function is given by \begin{align} L(x_1, x_2, \cdots, x_n; \theta_1,\theta_2)&=\frac{1}{(2 \pi)^{\frac{n}{2}} {\theta_2}^{\frac{n}{2}}} \exp \left({-\frac{1}{2 \theta_2} \sum_{i=1}^{n} (x_i-\theta_1)^2}\right). \end{align} Here again, it is easier to work with the log likelihood function \begin{align} \ln L(x_1, x_2, \cdots, x_n; \theta_1,\theta_2)&= -\frac{n}{2} \ln (2 \pi) -\frac{n}{2} \ln \theta_2 -\frac{1}{2 \theta_2} { \sum_{i=1}^{n} (x_i-\theta_1)^2}. \end{align} We take the derivatives with respect to $\theta_1$ and $\theta_2$ and set them to zero: \begin{align}%\label{} \frac{\partial }{\partial \theta_1} \ln L(x_1, x_2, \cdots, x_n; \theta_1,\theta_2) &=\frac{1}{\theta_2} \sum_{i=1}^{n} (x_i-\theta_1)=0 \\ \frac{\partial }{\partial \theta_2} \ln L(x_1, x_2, \cdots, x_n; \theta_1,\theta_2) &=-\frac{n}{2\theta_2}+\frac{1}{2\theta^2_2} \sum_{i=1}^{n}(x_i-\theta_1)^2=0. \end{align} By solving the above equations, we obtain the following maximum likelihood estimates for $\theta_1$ and $\theta_2$: \begin{align}%\label{} &\hat{\theta}_1=\frac{1}{n} \sum_{i=1}^{n} x_i,\\ &\hat{\theta}_2=\frac{1}{n} \sum_{i=1}^{n} (x_i-\hat{\theta}_1)^2. \end{align} We can write the MLE of $\theta_1$ and $\theta_2$ as random variables $\hat{\Theta}_1$ and $\hat{\Theta}_2$: \begin{align}%\label{} &\hat{\Theta}_1=\frac{1}{n} \sum_{i=1}^{n} X_i,\\ &\hat{\Theta}_2=\frac{1}{n} \sum_{i=1}^{n} (X_i-\hat{\Theta}_1)^2. \end{align} Note that $\hat{\Theta}_1$ is the sample mean, $\overline{X}$, and therefore it is an unbiased estimator of the mean. Here, $\hat{\Theta}_2$ is very close to the sample variance which we defined as \begin{align}%\label{} {S}^2=\frac{1}{n-1} \sum_{i=1}^n (X_i-\overline{X})^2. \end{align} In fact, \begin{align}%\label{} \hat{\Theta}_2=\frac{n-1}{n} {S}^2. \end{align} Since we already know that the sample variance is an unbiased estimator of the variance, we conclude that $\hat{\Theta}_2$ is a biased estimator of the variance: \begin{align}%\label{} E\hat{\Theta}_2=\frac{n-1}{n} \theta_2. \end{align} Nevertheless, the bias is very small here and it goes to zero as $n$ gets large. Note: Here, we caution that we cannot always find the maximum likelihood estimator by setting the derivative to zero. For example, if $\theta$ is an integer-valued parameter (such as the number of blue balls in Example 8.7), then we cannot use differentiation and we need to find the maximizing value in another way. Even if $\theta$ is a real-valued parameter, we cannot always find the MLE by setting the derivative to zero. For example, the maximum might be obtained at the endpoints of the acceptable ranges. We will see an example of such scenarios in the Solved Problems section (Section 8.2.5). The print version of the book is available through Amazon here.

http://mathhelpforum.com/calculus/128052-serious-series-problem.html

# Math Help - Serious Series Problem 1. ## Serious Series Problem What is the sum of: 1/1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ... + 1/(1+2+3+...+n) 2. Originally Posted by bearej50 What is the sum of: 1/1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ... + 1/(1+2+3+...+n) $\sum_{k=1}^{j}k=\frac{j(j+1)}{2}$. And so $\sum_{j=1}^{n}\frac{1}{\sum_{k=1}^{j}k}=\sum_{j=1} ^{n}\frac{2}{j(j+1)}=2\sum_{j=1}^{n}\left\{\frac{1 }{j}-\frac{1}{j+1}\right\}=2\left[1-\frac{1}{n+1}\right]$.

https://byjus.com/question-answer/for-what-value-of-n-frac-a-n-1-b-n-1-a-n-b/

Question # For what value of n, an+1+bn+1an+bn is the arithmetic mean of a and b? Open in App Solution ## Since, arithmetic mean of a and b is a+b2, therefore according to the given condition, an+1+bn+1an+bn=a+b2 ⇒ 2an+1+2bn+1=an+1+anb+abn+bn+1 ⇒ 2an+1−an+1−anb=abn+bn+1−2bn+1 ⇒ an+1−anb=abn−bn+1 ⇒ an(a−b)=bn(a−b) ⇒ an=bn [∵ a≠b] ⇒ (ab)n=1⇒(ab)n=(ab)0⇒ n=0 Hence, for n = 0, an+1+bn+1an+bn is the arithmetic mean of a and b. Suggest Corrections 14 Explore more

https://math.stackexchange.com/questions/1191691/solve-a-pde-with-feynman-kac-formula

# Solve a PDE with Feynman-Kac Formula So there is the following PDE given: $\frac{\partial}{\partial t}f(t,x) + rx\frac{\partial}{\partial x}f(t,x)+\frac{\sigma^2 x^2}{2}\frac{{\partial}^2}{\partial x^2}f(t,x) = rf(t,x)$ With boundary condition $f(T,x) = x^{\frac{2r}{\sigma^2}}$ Here $r$ and $\sigma$ are positive constants. From what I have learned, I the solution is from the boundary condition $f(t,x) =e^{-r(T-t)}E[x^{\frac{2r}{\sigma^2}}]$ So first I look fro the stochastic representation which I find as: $dX(t) = rdt + \sigma dW(t)$ with X(t) = x The solution is: $X(T) = x + r(T-t) + \sigma(W(T)-W(t))$ This is normally distributed with mean $x + r(T-t)$ and variance $\sigma \sqrt{T-t}$ Now from boundary condition I have $f(t,x) =e^{-r(T-t)}E[(x + r(T-t) + \sigma(W(T)-W(t)))^{\frac{2r}{\sigma^2}}]$ However I don't know if this method is correct. If it is correct, how should I calculate this expectation? By the way, I take this expectation under $Q$ martingale measure. Thanks Define $$f(x,t):=\mathbb E[e^{-r(T-t)}X_T^{\frac{2r}{\sigma^2}} \vert X_t=x]\,,$$ where the Ito-process, $\text dX_u = rX_u\text du + \sigma X_u\text dW_u\$ (with $X_t=x$), and the Weiner process, $W_u$, have been defined with respect to some suitable underlying filtered probability space (with filtration $\{\mathcal F_t\}_{t\geqslant 0}$). Now, by solving the SDE defining the process $X_u$, we obtain $$X_u=xe^{(r-\frac{1}{2}\sigma^2)(u-t)+\sigma (W_{u}-W_{t})}\ \ (\text{for }u\geqslant t).$$ Thus, by considering conditional expectations with respect to the sigma-algebra $\mathcal F_t$, $$\mathbb E[e^{-r(T-t)}X_T^{\frac{2r}{\sigma^2}} \vert \mathcal F_t] = e^{-r(T-t)}\mathbb E[X_T^{\frac{2r}{\sigma^2}} \vert \mathcal F_t] = e^{-r(T-t)}\mathbb E[(xe^{(r-\frac{1}{2}\sigma^2)(T-t)+\sigma (W_{T}-W_{t})})^{\frac{2r}{\sigma^2}} \vert \mathcal F_t] = e^{-r(T-t)}x^{\frac{2r}{\sigma^2}}e^{\frac{2r}{\sigma^2}(r-\frac{1}{2}\sigma^2)(T-t)}\mathbb E[(e^{\sigma (W_{T}-W_{t})})^{\frac{2r}{\sigma^2}} \vert \mathcal F_t] = e^{-r(T-t)}x^{\frac{2r}{\sigma^2}}e^{\frac{2r}{\sigma^2}(r-\frac{1}{2}\sigma^2)(T-t)}\mathbb E[e^{\frac{2r}{\sigma} (W_{T}-W_{t})}]\,,$$ where $\mathbb E[e^{\frac{2r}{\sigma} (W_{T}-W_{t})}\vert \mathcal F_t] = \mathbb E[e^{\frac{2r}{\sigma} (W_{T}-W_{t})}]$ since $(W_{T}-W_{t})$ is independent of $\mathcal F_t$. So, $$f(x,t) = x^{\frac{2r}{\sigma^2}}e^{-2r(T-t)}e^{2\frac{r^2}{\sigma^2}(T-t)}\mathbb E[e^{\frac{2r}{\sigma} (W_{T}-W_{t})}]\,.$$ We are almost done; we simply need to evaluate $\mathbb E[e^{\frac{2r}{\sigma} (W_{T}-W_{t})}]$ using the fact that $W_{T}-W_{t}\sim\mathcal N(0,T-t)$. Note that, $$\begin{eqnarray*} \mathbb E[e^{\frac{2r}{\sigma} (W_{T}-W_{t})}] &=& \frac{1}{\sqrt{2\pi(T-t)}}\int_{-\infty}^{\infty}e^{\frac{2r}{\sigma} w-\frac{1}{2}w^2/(T-t)}\text dw \\ && \\& = &\ldots \\ && \\ & = & e^{2\frac{r^2}{\sigma^2}(T-t)}\,. \end{eqnarray*}$$ Therefore, $$f(x,t) = x^{\frac{2r}{\sigma^2}}e^{-2r(T-t)}e^{4\frac{r^2}{\sigma^2}(T-t)}\,.$$ If you are curious, you can convince yourself that the solution above is correct by showing that it satisfies the given PDE: simply compute each of the partial derivatives and plug them in the PDE. • Hi, thanks. How did you come up with the Ito process? Why use exactly that one? Mar 16, 2015 at 18:24 • @Elekko, That is the form of Ito-process required by the Feynman-Kac formula (see en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula). Using the statement of the formula on that webpage, simply identify $\mu(x,t)$ and $\sigma^2(x,t)$ for your PDE and use these in the Ito-process on that webpage. – ki3i Mar 16, 2015 at 19:26

https://math.stackexchange.com/questions/2538366/for-what-values-of-alpha-does-int-0-infty-times-0-infty-frac

For what values of $\alpha$ does $\int_{(0, +\infty) \times (0, +\infty)} \frac{e^{-y} |\sin (x)|}{(1+xy)^{\alpha}}\: d(x, y)$ converge? For what values of $\alpha \in \mathbb R$ does $$\int_{(0, +\infty) \times (0, +\infty)} \frac{e^{-y} |\sin (x)|}{(1+xy)^{\alpha}}\: d(x, y)$$ converge? What I've tried: Using the substitution $(x, y) = (u, \frac{v}{u})$, we get $$\int_{(0, +\infty) \times (0, +\infty)} \frac{e^{-\frac{v}{u}} |\sin u|}{u (1+v)^{\alpha} }\: d(u, v).$$ Using the substitution $(x, y) = (\frac{u}{v}, v)$, we get $$\int_{(0, +\infty) \times (0, +\infty)} \frac{e^{-v} |\sin \frac{u}{v}|}{v (1+u)^{\alpha} }\: d(u, v).$$ How do I proceed?

https://giasutamtaiduc.com/unit-vector-formula.html

# Unit Vector Formula 5/5 - (1 bình chọn) Mục Lục ## Unit Vector Vectors are geometric entities that have magnitude and direction. Vectors have a starting point and a terminal point which represents the final position of the point. Various arithmetic operations can be applied to vectors such as addition, subtraction, and multiplication. A vector that has a magnitude of 1 is termed a unit vector. For example, vector v = (1, 3) is not a unit vector, because its magnitude is not equal to 1, i.e., |v| = √(12+32) ≠ 1. Any vector can become a unit vector when we divide it by the magnitude of the same given vector. A unit vector is also sometimes referred to as a direction vector. Let us learn more about the unit vector, its formula along with a few solved examples. ## What is Unit Vector? A unit vector is a vector that has the magnitude equal to 1. The unit vectors are denoted by the “cap” symbol ^. The length of unit vectors is 1. Unit vectors are generally used to denote the direction of a vector. A unit vector has the same direction as the given vector but has a magnitude of one unit; For a vector A; a unit vector is; ^A and ^A=(1/|A|)^A i, j, and k are the unit vectors in the directions of the x-axis, y-axis, and z-axis respectively in a 3-dimensional plane. i.e., • |i| = 1 • |j| = 1 • |k| = 1 ### Magnitude of a Vector The magnitude of a vector gives the numeric value for a given vector. A vector has both a direction and a magnitude. The magnitude of a vector formula summarises the individual measures of the vector along the x-axis, y-axis, and z-axis. The magnitude of a vector A is |A|. For a given vector with the direction along the x-axis, y-axis, and z-axis, the magnitude of the vector can be obtained by calculating the square root of the sum of the squares of its direction ratios. Let us understand it clearly from the below magnitude of a vector formula. For a vector A = ai + bj + ck its magnitude is: |A| = √(a2 + b2 + c2) For example, if A = 1i + 2j + 3k, then |A| = √(12+22+22) = √9 = 3 units. ## Unit Vector Notation Unit Vector is represented by the symbol ‘^’, which is called a cap or hat, such as ^a. It is given by ^a = a/|a| Where |a| is for norm or magnitude of vector a. It can be calculated using a unit vector formula or by using a calculator. ### Unit vector in three-dimension The unit vectors of i, j, and k are usually the unit vectors along the x-axis, y-axis, z-axis respectively. Every vector existing in the three-dimensional space can be expressed as a linear combination of these unit vectors. The dot product of two unit vectors is always a scalar quantity. On the other hand, the cross-product of two given unit vectors gives a third vector perpendicular (orthogonal) to both of them. ### Unit Normal Vector A ‘normal vector’ is a vector that is perpendicular to the surface at a defined point. It is also called “normal” to a surface containing the vector. The unit vector that is acquired after normalizing the normal vector is the unit normal vector, also known as the “unit normal.” For this, we divide a non-zero normal vector by its vector norm. ## Unit Vector Formula As vectors have both magnitude (value) and direction, they are shown with an arrow. In particular, ^a denotes a unit vector. If we want to find the unit vector of any vector, we divide it by the vector’s magnitude. Usually, the coordinates of x, y, z are used to represent any vector. A vector can be represented in two ways: 1. a = (x, y, z) using the brackets. 2. a = xi + yj +zk The formula for the magnitude of a vector is: |a|= (x2 + y2 + z2) The formula of unit vector in the direction of a given vector is: • Unit Vector = Vector/Vector’s magnitude ## How to Calculate the unit vector? To find a unit vector with the same direction as a given vector, simply divide the vector by its magnitude. For example, consider a vector v = (3, 4) which has a magnitude of |v|. If we divide each component of vector v by |v| to get the unit vector ^v which is in the same direction as v. |v| = (32 + 42) = 5 Thus, ^v = v / |v| = (3, 4) / 5 = (3/5, 4/5). How to represent vector in a bracket format? If a = (x, y, z), then the unit vector in the direction of a in bracket format is, ^a = a/|a| = (x,y,z)/(x2 + y2 + z2) = ( x/ (x2 + y2 + z2), y/(x2 + y2 + z2), z/(x2 + y2 + z2) ) How to represent vector in a unit vector component format? If a = xi + yj + zk is a vector then the unit vector in the direction of a in component format is, ^a = a/|a| = (xi + yj + zk)/ (x2 + y2 + z2) = x/(x2 + y2 + z2) . i + y/(x2 + y2 + z2) . j + z/(x2 + y2 + z2) . k Where x, y, z represent the value of the vector along the x- axis, y-axis, z-axis respectively and ^a is a unit vector, a is a vector, |a| is the magnitude of the vector, i, j, k are the directed unit vectors along the x, y, and z axes respectively. ## Application of Unit Vector Unit vectors specify the direction of a vector. Unit vectors can exist in both two and three-dimensional planes. Every vector can be represented with its unit vector in the form of its components. The unit vectors of a vector are directed along the axes. In the 3-d plane, the vector v will be identified by three perpendicular axes (x, y, and z-axis). In mathematical notations, the unit vector along the x-axis is represented by i. The unit vector along the y-axis is represented by j, and the unit vector along the z-axis is represented by k. The vector v can hence be written as: v = xi + yj + zk Electromagnetics deals with electric forces and magnetic forces. Here vectors come in handy to represent and perform calculations involving these forces. In day-to-day life, vectors can represent the velocity of an airplane or a train, where both the speed and the direction of movement are needed. ## Properties of Vectors The properties of vectors are helpful to gain a detailed understanding of vectors and also to perform numerous calculations involving vectors, A few important properties of vectors are listed here. • A . B = B. A • A × B B × A • i . i = j . j = k . k = 1 • i . j = j . k = k . i = 0 • i × i = j × j = k × k = 0 • i × j = k; j × k = i; k × i = j • j × i = –k; k × j = –i; i × k = –j • The dot product of two vectors is a scalar and lies in the plane of the two vectors. • The cross product of two vectors is a vector, which is perpendicular to the plane containing these two vectors. ## Examples on Unit Vector = A/|A| = (3i + 4j – 5k) / (5√2) Answer: Hence the unit vector is (3/5√2) i + (4/5√2) j – (5/5√2) k. Example 2: Find the vector of magnitude 8 units and in the direction of the vector i – 7j + 2k. Solution: Given vector A = i – 7j + 2k. |A| = √(12 + (-7)2 + (2)2) = √(1 + 49 + 4) = √54 = 3√6 The unit vector can be calculated using this below formula. = A/|A| = (i – 7j + 2k) / (3√6) The vector of magnitude 8 units = 8 × (i – 7j + 2k) / (3√6) Answer: Therefore the vector of magnitude 8 units = (4√6/9) · (i – 7j + 2k). Example 3: Find the unit vector parallel to the resultant of the vectors A = 2i – 3j + 4k and B = i + 5j – 2k. Solution: The resultant vector of the given two vectors is: A + B = (2i – 3j + 4k) + (-i + 5j – 2k) = i + 2j + 2k. Its magnitude is, |A + B| = √(12 + 22 + 22) = √(9) = 3 To find the unit vector parallel to the resultant of the given vectors, we divide the above resultant vector by its magnitude. Thus, the required unit vector is, (A + B) / |A + B| = (i + 2j + 2k) / 3 = 1/3 i + 2/3 j + 2/3 k Answer: 1/3 i + 2/3 j + 2/3 k. ## FAQs on Unit Vector ### What is the Definition of Unit Vector? A vector that has a magnitude of 1 is a unit vector. It is also known as a direction vector because it is generally used to denote the direction of a vector. The vectors i, j, k, are the unit vectors along the x-axis, y-axis, and z-axis respectively. ### How Do You Find the Unit Vector With the Same Direction as a Given Vector? To find a unit vector with the same direction as a given vector, we divide the vector by its magnitude. For example, consider a vector v = (1, 4) which has a magnitude of |v|. If we divide each component of vector v by |v| we will get the unit vector which is in the same direction as v. ### What is a Unit Vector Used For? Unit vectors are only used to specify the direction of a vector. Unit vectors exist in both two and three-dimensional planes. Every vector has a unit vector in the form of its components. The unit vectors of a vector are directed along the axes. ### What is a Unit Vector Formula? The unit vector is obtained by dividing the vector A with its magnitude |A|. The unit vector has the same direction coordinates as that of the given vector. = A/|A|. ### What is a Normal Unit Vector? A unit normal vector to a two-dimensional curve is a vector with magnitude 1 that is perpendicular to the curve at some point. Typically you look for a function that gives you all possible unit normal vectors of a given curve, not just one vector. ### How Do You Find the Unit Vector Perpendicular to Two Vectors? The cross-product of two non-parallel results in a vector that is a vector that is perpendicular to both of them. So, for the given two vectors x and y, we know that, x × y will be a vector that is perpendicular to both x and y. Further, to find the unit vector of this resultant vector, we divide it by its magnitude. i.e., (x × y) / |x × y|, this would give the unit vector that is perpendicular to the given two vectors. ### When are the Two Vectors said to be Parallel Vectors? Two or more vectors are parallel if they are moving in the same direction. Also, the cross-product of parallel vectors is always zero. ## Unit Vector Symbol Unit Vector is represented by the symbol ‘^’, which is called a cap or hat, such as: â. It is given by Unit vectors are usually determined to form the base of a vector space. Every vector in the space can be expressed as a linear combination of unit vectors. The dot products of two unit vectors is a scalar quantity whereas the cross product of two arbitrary unit vectors results in third vector orthogonal to both of them. What is the unit normal vector? The normal vector is a vector which is perpendicular to the surface at a given point. It is also called “normal,” to a surface is a vector. When normals are estimated on closed surfaces, the normal pointing towards the interior of the surface and outward-pointing normal are usually discovered. The unit vector acquired by normalizing the normal vector is the unit normal vector, also known as the “unit normal.”Here, we divide a nonzero normal vector by its vector norm. ## Unit Vector Formula As explained above vectors have both magnitude (Value) and a direction. They are shown with an arrow. The above is a unit vector formula. How to find the unit vector? To find a unit vector with the same direction as a given vector, we divide the vector by its magnitude. For example, consider a vector v = (1, 4) which has a magnitude of |v|. If we divide each component of vector v by |v| we will get the unit vector uv which is in the same direction as v. How to represent Vector in a bracket format? ### Unit Vector Example Here is an example based on the unit vector. Observe and follow each step and solve problems based on it. Question 1: ### Magnitude of Unit Vector In order to calculate the numeric value of a given vector, the magnitude of the vector formula is used. The magnitude of a vector A⃗ is |A|. The magnitude of a vector can be identified by calculating the square roots of the sum of squares of its direction vectors. The magnitude of a vector formula is given by: will be read as A cap. For a unit vector u in the same direction as vector v, we divide the vector by its magnitude ### Representation of a Vector There are two ways in which a vector can be represented- Solved Examples Related to Unit Vector Unit Tangent Vector ### Solved Examples Example 1: Suppose, There is a vector a= (1,0). To find the magnitude of the given vector, we use the formula, |a|= √(x2+y2+z2) ……(1) Using the magnitude of unit vector formula (1), |a|= √ (12+0) =√1=1 Therefore, a is a unit vector. Example 2: Suppose, There is a vector b= (2, 3). To find the magnitude of the given vector, we use the formula, |b|= √(x2+y2+z2) ……(1) Using the magnitude of unit vector formula (1), |b|= √ (22+32) = √13 Now, √13≠1. Hence, from the above calculations, we can conclude that, a is not a unit vector. ### Formula To Find Unit Vector â = a/|a|…….(2) = (x,y,z)/√(x2+y2+z2) For example, a= (12,4,3) Then, using (1), |a|= √ ((122) +(42) +(32)) = √144+16+9 = √169=13 Substituting, |a| in vector unit formula (2), â = (12,4,3)/13 = ((12/13), (4/13), (3/13)) ### Sample Problems Problem 1. Find the unit vector of 2i + 4j + 5k. Solution: Problem 2. Find the unit vector of 3i + 4j + 5k. Solution: Problem 3. Find the unit vector of i + 2j + 2k. Solution: Problem 4. Find the unit vector of the resultant vector of i + 3j +5k and -j – 3k. Solution: Problem 5. Find the unit vector of 4i + 4j. Solution: Mọi chi tiết liên hệ với chúng tôi : TRUNG TÂM GIA SƯ TÂM TÀI ĐỨC Số điện thoại tư vấn cho Phụ Huynh : Điện Thoại : 091 62 65 673 Các số điện thoại tư vấn cho Gia sư : Điện thoại : 0946433647 hoặc 0908 290 601

https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-15-differentiation-in-several-variables-15-3-partial-derivatives-exercises-page-781/43

Calculus (3rd Edition) $$g_u(1,2)= \frac{1}{3}+\ln 3 .$$ Since $g(u,v)=u\ln (u+v)$, then by the product rule, we have $$g_u=\ln (u+v)+u\frac{1}{u+v}=\ln (u+v)+\frac{u}{u+v}.$$ Hence, we get $$g_u(1,2)=\ln (1+3)+\frac{1}{1+2}=\frac{1}{3}+\ln 3 .$$

https://www.coursehero.com/file/88556361/Module-7-APCalcBC-7-6-Improper-Integralspptx/

Module 7_ APCalcBC 7-6 Improper Integrals.pptx - 7.6 Improper Integrals 1\/22 Do Now Evaluate 2 2x 4 lim 3 x\u00ae \u00a5 3x x 2 HW Review Improper Integrals \u2022 # Module 7_ APCalcBC 7-6 Improper Integrals.pptx - 7.6... • 14 This preview shows page 1 - 14 out of 14 pages. 7.6 Improper Integrals 1/22 Do Now Evaluate lim x ®¥ 2 x 2 - 4 3 x 3 + x - 2 HW Review Improper Integrals Areas that are unbounded are represented by improper integrals An integral is improper if The interval of integration may be infinite (bound to infinity) The integrand may tend to infinity (vertical asymptote in the bounds) Improper integral Assume f(x) is integrable over [a,b] for all b>a. The improper integral of f(x) is defined as The improper integral converges if the limit exists (and is finite) and diverges if the limit does not exist (or is infinite) f ( x ) dx a ¥ ò = lim R ®¥ f ( x ) dx a R ò Ex Evaluate dx x 3 2 ¥ ò Ex Determine whether converges or not dx x - ¥ - 1 ò The p-integral For a > 0, if P > 1 The integral diverges if P <= 1 dx x P a ¥ ò = a 1 - p p - 1 Ex Evaluate xe - x dx 0 ¥ ò Comparing Integrals

http://tasks.illustrativemathematics.org/content-standards/HSS/ID/C/7/tasks/1028

Engage your students with effective distance learning resources. ACCESS RESOURCES>> Alignments to Content Standards: S-ID.C.7 Medhavi suspects that there is a relationship between the number of text messages high school students send and their academic achievement. To explore this, she asks a random sample of 52 students at her school how many text messages they sent yesterday and what their grade point average (GPA) was during the most recent marking period. Her data are summarized in the scatter plot below. The line of best fit is also shown. The equation of the line of best fit is $\widehat{GPA}=3.8 - 0.005(\text{Texts sent})$. Interpret the quantities $-0.005$ and $3.8$ in the context of these data. IM Commentary The purpose of this task is to assess ability to interpret the slope and intercept of the line of best fit in context. There are two common errors that students make when interpreting the slope. Students may not make it clear that the slope is the predicted change (not necessarily an actual change) in GPA associated with an increase of 1 in number of text messages sent. They also often do not clearly communicate that the slope describes change You might want to point out that it is not always reasonable to interpret the intercept as the predicted y value when x = 0, as this often involves extrapolation far beyond the range of the x values in the data set. In this example, however, it is appropriate because there are observations with x = 0 in the data set. You can also point out that the interpretation of the slope and intercept represents a generalization from the sample of 52 students to the population of all students at the school. This is appropriate because the sample was a random sample of students from the school. Although this task is short and looks simple, some of the points brought out in this task are subtle. It might be a good strategy to engage in a whole class discussion of the correct interpretations. Solution Interpretation of the slope: For students at this school, the predicted GPA decreases by 0.005 for each additional text messaage sent OR GPA desreases by 0.005, on average, for each additional text message sent. Interpretation of intercept: The model predicts that students at this school who send no text messages have, on average, a GPA of 3.8.

https://socratic.org/questions/solve-the-equation-4

# Solve the equation? ## $\sqrt{{x}^{2} + 4 x - 21} + \sqrt{{x}^{2} - x - 6} = \sqrt{6 {x}^{2} - 5 x - 39}$ Jun 25, 2018 $x = 3$ #### Explanation: $\sqrt{{x}^{2} + 4 x - 21} + \sqrt{{x}^{2} - x - 6} = \sqrt{6 {x}^{2} - 5 x - 39}$ ${\left(\sqrt{{x}^{2} + 4 x - 21} + \sqrt{{x}^{2} - x - 6}\right)}^{2} = {\left(\sqrt{6 {x}^{2} - 5 x - 39}\right)}^{2}$ ${x}^{2} + 4 x - 21 + 2 \cdot \sqrt{{x}^{2} + 4 x - 21} \cdot \sqrt{{x}^{2} - x - 6} + {x}^{2} - x - 6 = 6 {x}^{2} - 5 x - 39$ $2 {x}^{2} + 3 x - 27 + 2 \sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)} = 6 {x}^{2} - 5 x - 39$ $2 \sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)} = 4 {x}^{2} - 8 x - 12$ $\sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)} = 2 {x}^{2} - 4 x - 6$ ${\left(\sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)}\right)}^{2} = {\left(2 {x}^{2} - 4 x - 6\right)}^{2}$ $\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right) = {\left(2\right)}^{2} {\left({x}^{2} - 2 x - 3\right)}^{2}$ $\left(x + 7\right) \left(x - 3\right) \left(x - 3\right) \left(x + 2\right) = 4 {\left(\left(x - 3\right) \left(x + 1\right)\right)}^{2}$ $\left(x + 7\right) {\left(x - 3\right)}^{2} \left(x + 2\right) = 4 {\left(x - 3\right)}^{2} {\left(x + 1\right)}^{2}$ ${\left(x - 3\right)}^{2} \left(\left(x + 7\right) \left(x + 2\right) - 4 {\left(x + 1\right)}^{2}\right) = 0$ ${\left(x - 3\right)}^{2} \left({x}^{2} + 9 x + 14 - 4 \left({x}^{2} + 2 x + 1\right)\right) = 0$ ${\left(x - 3\right)}^{2} \left({x}^{2} + 9 x + 14 - 4 {x}^{2} - 8 x - 4\right) = 0$ ${\left(x - 3\right)}^{2} \left(- 3 {x}^{2} + x + 10\right) = 0$ $- {\left(x - 3\right)}^{2} \left(3 {x}^{2} - x - 10\right) = 0$ $- {\left(x - 3\right)}^{2} \left(3 x + 5\right) \left(x - 2\right) = 0$ $x = 3$ or $x = - \frac{5}{3}$ or $x = 2$ However, as @Mark D points out, all the solutions but $x = 3$ give negative numbers within the square roots and so only $x = 3$ is valid: $\sqrt{{x}^{2} - x - 6}$ $\sqrt{{2}^{2} - 2 - 6} \implies \sqrt{4 - 2 - 6} \implies \sqrt{- 4}$ $\sqrt{{\left(- \frac{5}{3}\right)}^{2} + \frac{5}{3} - 6} \implies \sqrt{\frac{25}{9} + \frac{15}{9} - \frac{54}{9}} \implies \sqrt{- \frac{14}{9}}$

https://math.stackexchange.com/questions/3228923/binomial-coefficients-with-variable-in-exponent

Binomial coefficients with variable in exponent I need to calculate the coefficient of a specific term in a binomial, but how do I do that if the exponent has a variable in it? For example: Find the coefficient of $$x^n$$ in the expansion of $$(4x + 5x^2)^{7n}$$ or Find the coefficient of $$x^5$$ in the expansion of $$(3 + 2x^2)^{5n}$$ Note that these examples are not a homework problems; I do not need just the answer. I am trying to learn how to solve very similar problems because my text does not explain (and I can't figure it out). Looking at your first example: $$(4x + 5x^2)^{7n} = \sum_{k=0}^{7n}\binom{7n}{k}4^kx^k5^{7n-k}x^{2(7n-k)}$$ So, for which of the $$k$$ we have $$k+2(7n-k)=n$$? Solving this for $$k$$ we get $$k=13n$$ which (for $$n>0$$) is not included in the sum (so the coefficient of $$x^n$$ is zero). The second example is even easier: For any integer $$n$$, the exponents of $$x$$ in $$(3 + 2x^2)^{5n}$$ all are even... In general, you have to write down the binomial formula (as somthing like $$\sum_{k=0}^{f(n)}\ldots$$) for your term and solve the desired equation for the exponents of $$x$$ for $$k$$. Then, knowing all these $$k$$, you can evaluate the sum for just these $$k$$ (which is probably just a single term) with $$x=1$$ to find the desired coefficient.

https://www.shaalaa.com/question-bank-solutions/in-a-town-there-are-94500-people-2-9-of-them-are-foreigners-6400-are-immigrants-and-the-rest-are-natives-how-many-are-natives-number-system-entrance-exam_106999

In a town, there are 94500 people. 2/9 of them are foreigners, 6400 are immigrants and the rest are natives. How many are natives? - Mathematics MCQ In a town, there are 94500 people. 2/9 of them are foreigners, 6400 are immigrants and the rest are natives. How many are natives? • 67100 • 27400 • 77600 • 88100 Solution 67100 Explanation: Let number of natives be x. According to the question, 2/9xx94500+6400+"x"=94500 21000 + 6400 + x = 94500 x = 94500 - 27400 = 67100 Concept: Number System (Entrance Exam) Is there an error in this question or solution?

https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/section-exercises-34/

## Section Exercises 1. With what kind of exponential model would half-life be associated? What role does half-life play in these models? 2. What is carbon dating? Why does it work? Give an example in which carbon dating would be useful. 3. With what kind of exponential model would doubling time be associated? What role does doubling time play in these models? 4. Define Newton’s Law of Cooling. Then name at least three real-world situations where Newton’s Law of Cooling would be applied. 5. What is an order of magnitude? Why are orders of magnitude useful? Give an example to explain. 6. The temperature of an object in degrees Fahrenheit after t minutes is represented by the equation $T\left(t\right)=68{e}^{-0.0174t}+72$. To the nearest degree, what is the temperature of the object after one and a half hours? For the following exercises, use the logistic growth model $f\left(x\right)=\frac{150}{1+8{e}^{-2x}}$. 7. Find and interpret $f\left(0\right)$. Round to the nearest tenth. 8. Find and interpret $f\left(4\right)$. Round to the nearest tenth. 9. Find the carrying capacity. 10. Graph the model. 11. Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data. x f (x) –2 0.694 –1 0.833 0 1 1 1.2 2 1.44 3 1.728 4 2.074 5 2.488 12. Rewrite $f\left(x\right)=1.68{\left(0.65\right)}^{x}$ as an exponential equation with base e to five significant digits. For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that is linear, exponential, or logarithmic. 13. x f (x) 1 2 2 4.079 3 5.296 4 6.159 5 6.828 6 7.375 7 7.838 8 8.238 9 8.592 10 8.908 14. x f(x) 1 2.4 2 2.88 3 3.456 4 4.147 5 4.977 6 5.972 7 7.166 8 8.6 9 10.32 10 12.383 15. x f(x) 4 9.429 5 9.972 6 10.415 7 10.79 8 11.115 9 11.401 10 11.657 11 11.889 12 12.101 13 12.295 16. x f(x) 1.25 5.75 2.25 8.75 3.56 12.68 4.2 14.6 5.65 18.95 6.75 22.25 7.25 23.75 8.6 27.8 9.25 29.75 10.5 33.5 For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in t years is modeled by the equation $P\left(t\right)=\frac{1000}{1+9{e}^{-0.6t}}$. 17. Graph the function. 18. What is the initial population of fish? 19. To the nearest tenth, what is the doubling time for the fish population? 20. To the nearest whole number, what will the fish population be after 2 years? 21. To the nearest tenth, how long will it take for the population to reach 900? 22. What is the carrying capacity for the fish population? Justify your answer using the graph of P. 23. A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay? 24. The formula for an increasing population is given by $P\left(t\right)={P}_{0}{e}^{rt}$ where ${P}_{0}$ is the initial population and > 0. Derive a general formula for the time t it takes for the population to increase by a factor of M. 25. Recall the formula for calculating the magnitude of an earthquake, $M=\frac{2}{3}\mathrm{log}\left(\frac{S}{{S}_{0}}\right)$. Show each step for solving this equation algebraically for the seismic moment S. 26. What is the y-intercept of the logistic growth model $y=\frac{c}{1+a{e}^{-rx}}$? Show the steps for calculation. What does this point tell us about the population? 27. Prove that ${b}^{x}={e}^{x\mathrm{ln}\left(b\right)}$ for positive $b\ne 1$. For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. 28. To the nearest hour, what is the half-life of the drug? 29. Write an exponential model representing the amount of the drug remaining in the patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram. 30. Using the model found in the previous exercise, find $f\left(10\right)$ and interpret the result. Round to the nearest hundredth. For the following exercises, use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. 31. To the nearest day, how long will it take for half of the Iodine-125 to decay? 32. Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram. 33. A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance? 34. The half-life of Radium-226 is 1590 years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits. 35. The half-life of Erbium-165 is 10.4 hours. What is the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits. 36. A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 5730 years.) 37. A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours? For the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes. 38. To the nearest whole number, what was the initial population in the culture? 39. Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double? For the following exercises, use this scenario: A pot of boiling soup with an internal temperature of 100º Fahrenheit was taken off the stove to cool in a 69ºF room. After fifteen minutes, the internal temperature of the soup was 95ºF. 40. Use Newton’s Law of Cooling to write a formula that models this situation. 41. To the nearest minute, how long will it take the soup to cool to 80ºF? 42. To the nearest degree, what will the temperature be after 2 and a half hours? For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of 165ºF and is allowed to cool in a 75ºF room. After half an hour, the internal temperature of the turkey is 145ºF. 43. Write a formula that models this situation. 44. To the nearest degree, what will the temperature be after 50 minutes? 45. To the nearest minute, how long will it take the turkey to cool to 110ºF? For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth. 46. 47. 48. Plot each set of approximate values of intensity of sounds on a logarithmic scale: Whisper: ${10}^{-10} \frac{W}{{m}^{2}}$, Vacuum: ${10}^{-4}\frac{W}{{m}^{2}}$, Jet: ${10}^{2} \frac{W}{{m}^{2}}$ 49. Recall the formula for calculating the magnitude of an earthquake, $M=\frac{2}{3}\mathrm{log}\left(\frac{S}{{S}_{0}}\right)$. One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth. For the following exercises, use this scenario: The equation $N\left(t\right)=\frac{500}{1+49{e}^{-0.7t}}$ models the number of people in a town who have heard a rumor after t days. 50. How many people started the rumor? 51. To the nearest whole number, how many people will have heard the rumor after 3 days? 52. As t increases without bound, what value does N(t) approach? Interpret your answer. For the following exercise, choose the correct answer choice. 53. A doctor and injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation? A. $f\left(t\right)=13{\left(0.0805\right)}^{t}$ B. $f\left(t\right)=13{e}^{0.9195t}$ C. $f\left(t\right)=13{e}^{\left(-0.0839t\right)}$ D. $f\left(t\right)=\frac{4.75}{1+13{e}^{-0.83925t}}$

https://naturale0.github.io/nonparametric/asymptotics_is_strange

Asymptotics is strange... It is a useful trick to “flip” the denominator into numerator when it comes to proving asymptotic properties of errors. Here’s how to do so. Consider a form $\frac{1}{b_0 + b_1 h + o(h)},$ where $h \to 0$ and $b_i$’s are non-zero constants. Then \begin{aligned} &\frac{1}{b_0 + b_1 h + o(h)} \\ &= \frac{1}{b_0\left(1 + \frac{b_1}{b_0} h + o(h)\right)} \\ &= \frac{1}{b_0} \left(1 - \frac{b_1}{b_0} h + o(h)\right). \end{aligned} The trick is not only easy to use but easy to understand as well. Although there isn’t any technically difficult part, staring at the result I could not help but say this. Asymptotics is strange…

https://math.stackexchange.com/questions/2834183/minimum-generator-set-for-matrix-algebras

# Minimum generator set for matrix algebras Let $A\subseteq M_n(\mathbb{R})$ a finite set, where $M_n(\mathbb{R})$ is the set of all $n\times n$ matrices. Let $S(A)$ the generated algebra by $A$. What is the minimun number of elements of $A$ in order that $S(A)=M_n(\mathbb{R})$? In other words: what is the minimum $k$ such that there are $A_1,...,A_k$ elements of $M_n(\mathbb{R})$ with the property that every element of $M_n(\mathbb{R})$ is a polynomial on $A_1,...,A_n$? I know that there is a lot of articles about the exact number of elements of $A$, but i believe this is a simpler question. I can do it with $2$ matrices. One matrix $X$ with all entries $0$ except for one (say the top left) where we have a $1$, and one order-$n$ cyclic permutation matrix $Y$. Any matrix with a single entry $1$ and the rest $0$ may be written as a product $Y^aXY^b$ for natural numbers $a,b$. Any matrix in $M_n(\Bbb R)$ may be written as a linear combination of those.

https://mc-stan.org/docs/2_23/functions-reference/multivariate-gaussian-process-distribution.html

This is an old version, view current version. ## 21.4 Multivariate Gaussian Process Distribution ### 21.4.1 Probability Density Function If $$K,N \in \mathbb{N}$$, $$\Sigma \in \mathbb{R}^{N \times N}$$ is symmetric, positive definite kernel matrix and $$w \in \mathbb{R}^{K}$$ is a vector of positive inverse scales, then for $$y \in \mathbb{R}^{K \times N}$$, $\text{MultiGP}(y|\Sigma,w) = \prod_{i=1}^{K} \text{MultiNormal}(y_i|0,w_i^{-1} \Sigma),$ where $$y_i$$ is the $$i$$th row of $$y$$. This is used to efficiently handle Gaussian Processes with multi-variate outputs where only the output dimensions share a kernel function but vary based on their scale. Note that this function does not take into account the mean prediction. ### 21.4.2 Sampling Statement y ~ multi_gp(Sigma, w) Increment target log probability density with multi_gp_lpdf(y | Sigma, w) dropping constant additive terms. ### 21.4.3 Stan Functions real multi_gp_lpdf(matrix y | matrix Sigma, vector w) The log of the multivariate GP density of matrix y given kernel matrix Sigma and inverses scales w

https://mdtpmodules.org/linr/linr-2/linr-1-lesson-4-writing-the-equation-of-a-line/linr-1-lesson-4-try-this-convert-equations-to-standard-form/linr-1-lesson-4-try-this-converting-between-slope-intercept-and-standard-form-solutions/

# LINR 1 | Lesson 4 | Try This! (Converting between Slope-Intercept and Standard Form Solutions) Investigate this idea to develop a process for isolating a variable. 1. Use the equation $$3x-2y=6$$. • Substitute the value $$x=4$$ into the equation $$3x-2y=6$$.  Isolate the variable $$y$$ by solving the equation for $$y$$. $3x-2y=6$ $3(4)-2y=6$ $12-2y=6$ $12-12-2y=6-12$ $-2y=-6$ $\frac{-2y}{-2}=\frac{-6}{-2}$ $y=3$ • Substitute the value $$x=-6$$ into the equation $$3x-2y=6$$.  Isolate the variable $$y$$ by solving the equation for $$y$$. $3x-2y=6$ $3(-6)-2y=6$ $-18-2y=6$ $-18+18-2y=6+18$ $-2y=24$ $\frac{-2y}{-2}=\frac{24}{-2}$ $y=-12$ • Now solve the equation $$3x-2y=6$$ for $$y$$ without knowing a value for $$x$$.  Leave your answer in terms of $$x$$ which means that “$$x$$” will remain in your final equation. $3x-2y=6$ $3x-3x-2y=6-3x$ $-2y=-3x+6$ $\frac{-2y}{-2}=\frac{-3x+6}{-2}$ $y=\frac{3}{2}x-3$ • What is the slope of the line $$3x-2y=6$$?  The slope = $$\frac{3}{2}$$ • What is the $$y$$-intercept of the line $$3x-2y=6$$?  The $$y$$-intercept is $$(0,-3)$$ 2. Solve the equation $$4x+2y=24$$ for “$$y$$” in terms of “$$x$$” and identify the slope and the $$y$$-intercept. $4x+2y=24$ $4x-4x+2y=-4x+24$ $\frac{2y}{2}=\frac{-4x+24}{2}$ $y=-2x+12$

https://blog.zilin.one/2015/02/25/an-upper-bound-on-stirling-number-of-the-second-kind/

# An Upper Bound on Stirling Number of the Second Kind We shall show an upper bound on the Stirling number of the second kind, a byproduct of a homework exercise of Probabilistic Combinatorics offered by Prof. Tom Bohman. Definition. A Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set of $n$ objects into $k$ non-empty subsets and is denoted by $S(n,k)$. Proposition. For all $n, k$, we have $$S(n,k) \leq \frac{k^n}{k!}\left(1-(1-1/m)^k\right)^m.$$ Proof. Consider a random bipartite graph with partite sets $U:=[n], V:=[k]$. For each vertex $u\in U$, it (independently) connects to exactly one of the vertices in $V$ uniformly at random. Suppose $X$ is the set of non-isolated vertices in $V$. It is easy to see that $$\operatorname{Pr}\left(X=V\right) = \frac{\text{number of surjections from }U\text{ to }V}{k^n} = \frac{k!S(n,k)}{k^n}.$$ On the other hand, we claim that for any $\emptyset \neq A \subset [k]$ and $i \in [k]\setminus A$, $$\operatorname{Pr}\left(i\in X \mid A\subset X\right) \leq \operatorname{Pr}\left(i\in X\right).$$ Note that the claim is equivalent to $$\operatorname{Pr}\left(A\subset X \mid i\notin X\right) \geq \operatorname{Pr}\left(A\subset X\right).$$ Consider the same random bipartite graph with $V$ replaced by $V':=[k]\setminus \{i\}$ and let $X'$ be the set of non-isolated vertices in $V'$. The claim is justified since $$\operatorname{Pr}\left(A\subset X\mid i\notin X\right) = \operatorname{Pr}\left(A\subset X'\right) \geq \operatorname{Pr}\left(A\subset X\right).$$ Set $A:=[i-1]$ in above for $i = 2, \ldots, k$. Using the multiplication rule with telescoping the conditional probability, we obtain \begin{aligned}\operatorname{Pr}\left(X=V\right) =& \operatorname{Pr}\left(1\in X\right)\operatorname{Pr}\left(2\in X \mid [1]\subset X\right) \\ & \ldots \operatorname{Pr}\left(k\in X\mid [k-1]\subset X\right)\\ \leq & \operatorname{Pr}\left(1\in X\right)\operatorname{Pr}\left(2\in X\right)\ldots\operatorname{Pr}\left(k\in X\right) \\ = & \left(1-(1-1/m)^k\right)^m.\end{aligned}

https://deeplearningmath.org/logistic-regression-type-neural-networks.html

# 2 Logistic Regression Type Neural Networks Learning outcomes from this chapter • Logistic regression view as a shallow Neural Network • Maximum Likelihood, loss function, cross-entropy • Softmax regression/ multinomial regression model as a Multiclass Perceptron. • Optimisation procedure: gradient descent, stochastic gradient descent, Mini-Batches • Understand the forward pass and backpropogration step • Implementation from first principles ## 2.1 Logistic regression view as a shallow Neural Network ### 2.1.1 Sigmoid function The sigmoid function $$\sigma(\cdot)$$, also known as the logistic function, is defined as follows: $\forall z\in\mathbb{R},\quad \sigma(z)=\frac{1}{1+e^{-z}}\in]0,1[$ Figure 2.1: Sigmoid function z <- seq(-5, 5, 0.01) sigma = 1 / (1 + exp(-z)) plot(sigma~z,type="l",ylab=expression(sigma(z))) ### 2.1.2 Logistic regression The logistic regression is a probabilistic model that aims to predict the probability that the outcome variable $$y$$ is 1. It is defined by assuming that $$y|x;\theta\sim\textrm{Bernoulli}(\phi)$$. Then, the logistic regression is defined by applying the soft sigmoid function to the linear predictor $$\theta^Tx$$: $\phi=h_{\theta}(x)=p(y=1|x;\theta)=\frac{1}{1+\exp(-\theta^Tx)}=\sigma(\theta^Tx)$ The logistic regression is also presented: $\textrm{Logit}[h_{\theta}(x)]=logit[p(y=1|x;\theta)]=\theta^Tx$ where $$\textrm{Logit}(p)=log\left(\frac{p}{1-p}\right)$$. • $$x=(x_0,\dots,x_d)^T$$ represent a vector of $$d+1$$ features/predictors and by convention $$x_0=1$$ • $$\theta=(\theta_0,\dots,\theta_d)^T$$ is the vector of parameter related to the features $$x$$ • $$\theta_0$$ is called the intercept by the statistician and named bias by the computer scientist (noted generally $$b$$) ### 2.1.3 Logistic regression for classification Let’s play with a simple example with 10 points, and two classes (red and blue) Figure 2.2: Classify red and blue points clr1 <- c(rgb(1,0,0,1),rgb(0,0,1,1)) clr2 <- c(rgb(1,0,0,.2),rgb(0,0,1,.2)) x <- c(.4,.55,.65,.9,.1,.35,.5,.15,.2,.85) y <- c(.85,.95,.8,.87,.5,.55,.5,.2,.1,.3) z <- c(1,1,1,1,1,0,0,1,0,0) df <- data.frame(x,y,z) plot(x,y,pch=19,cex=2,col=clr1[z+1]) In order to classify the points, we run a logistic regression to get predictions model <- glm(z~x+y,data=df,family=binomial) #summary(model) Then, we use the fitted model to define our classifier which is defined as attributed the class that is the most likely. pred_model <- function(x,y){ predict(model,newdata=data.frame(x=x, y=y),type="response")>.5 } Using our decision rule, we can visualise the produced partition of the space. Figure 2.3: Partition using the logistic model x_grid<-seq(0,1,length=101) y_grid<-seq(0,1,length=101) z_grid <- outer(x_grid,y_grid,pred_model) image(x_grid,y_grid,z_grid,col=clr2) points(x,y,pch=19,cex=2,col=clr1[z+1]) ### 2.1.4 Likelihood of the logistic model The maximum likelihood estimation procedure is generally used to estimate the parameters of the models $$\theta_0,\ldots,\theta_d$$. $p(y|x;\theta) = \begin{cases} h_\theta(x) & \text{if } y = 1, \text{ and} \\ 1 - h_\theta(x) & \text{otherwise}. \end{cases}$ which could be written as $p(y|x;\theta) = h_\theta(x)^y(1-h_\theta(x))^{1-y},$ Consider now the observation of $$m$$ training samples denoted by $$\left\{(x^{(1)},y^{(1)}),\ldots,(x^{(m)},y^{(m)})\right\}$$ as i.i.d. observations from the logistic model. The likelihood is $\begin{eqnarray*} L(\theta)&=&\prod_{i=1}^mp(y^{(i)}|x^{(i)};\theta)\\ &=&\prod_{i=1}^m h_\theta(x^{(i)})^y(1-h_\theta(x^{(i)}))^{1-y} \end{eqnarray*}$ Then, the following log likelihood is maximized to the estimates of $$\theta$$: $\ell(\theta)=\textrm{log }L(\theta)=\sum_{i=1}^m\left[y^{(i)}\log{h_\theta(x^{(i)})}+(1-y^{(i)})\log{(1-h_\theta(x^{(i)}))}\right]$ ### 2.1.5 Shallow Neural Network The logistic model can ve viewed as a shallow Neural Network. This figure used here the same notation as the regression logistic model presented by the statistical point of view. However, in the following we will adopt the notation used the most frequently in deep learning framework. In this figure, $$z=w^Tz+b=w_1x_1+\ldots+w_dx_d+b$$ is the linear combination of the $$d$$ features/predictors and $$a=\sigma(Z)$$ is called the activation function which is the non-linear part of the Neural Network to get a close prediction $$\hat{y}\approx y$$. Remark: In the sequel, we will adopt the following notation: • $$x=(x_1,\dots,x_d)^T\in \Re^d$$ representz a vector of $$d$$ features/predictors • $$w=(w_1,\dots,w_d)^T$$ is the vector of weight related to the features $$x$$ • $$b$$ is called the biais • We consider the observations of $$m$$ training samples denoted by $$\left\{(x^{(1)},y^{(1)}),\ldots,(x^{(m)},y^{(m)})\right\}$$ ### 2.1.6 Entropy, Cross-entropy and Kullback-Leibler Let’s first talk about the cross-entropy which is widely used as loss function for classification purpose. Cross Entropy (CE) is related to the entropy and the kullback-Leibler risk. The entropy of a discrete probability distribution $$p=(p_1,\ldots,p_n)$$ is defined as $H(p)=H(p_1,\ldots,p_n)=-\sum_{i=1}^np_i\log p_i$ which is a measurement of the disorder or randomness of a system’’. Kullback and Leibler known also as KL divergence quantifies how similar a probability distribution $$p$$ is to a candidate distribution $$q$$. $KL(p;q)=-\sum_{i=1}^np_i\log \frac{p_i}{q_i}$ Note that the $$KL$$ divergence is not a distance measure as $$KL(p;q)\ne KL(q;p)$$. $$KL$$ is non-negative and zero if and only if $$p_i = q_i$$ for all $$i$$. One can easily show that $KL(p;q)=\underbrace{\sum_{i=1}^np_i\log \frac{1}{q_i}}_{\textrm{cross entropy}}-H(p)$ where the first term of the right part is the cross entropy: $CE(p,q)=\sum_{i=1}^np_i\log \frac{1}{q_i}=-\sum_{i=1}^np_i\log q_i$ And we have the relation $CE(p,q)=H(p)+KL(p;q)$ Thus, the cross entropy can be interpreted as the uncertainty implicit in $$H(p)$$ plus the likelihood that the distribution $$p$$ could have be generated by the distribution $$q$$. ### 2.1.7 Mathematical expression of the Neural Network: For one example $$x^{(i)}$$, the ouput of this Neural Network is given by: $\hat{y}^{(i)}=\underbrace{\sigma(w^Tx^{(i)}+b)}_{\underbrace{a^{(i)}}_\textrm{activation function}},$ where $$\sigma(\cdot)$$ is the sigmoid function. We aim to get the weight $$w$$ and the biais $$b$$ such that $$\hat{y}^{(i)}\approx y^{(i)}$$. The loss function used for this network is the cross-entropy which is defined for one sample $$(x,y)$$: $\mathcal L(\hat{y},y)=-(y\log \hat{y} + (1-y)\log (1-\hat{y}))$ Then the Cost function for the entire training data set is: $J(w,b) = \frac{1}{m} \sum_{i=1}^m \mathcal{L}(\hat{y}^{(i)}, y^{(i)})$ Further, it is easy to see the connection with the log-likelihood function of the logistic model: $\begin{eqnarray*} J(w,b) &=& \frac{1}{m} \sum_{i=1}^m \mathcal{L}(\hat{y}^{(i)}, y^{(i)})\\ &=&-\frac{1}{m}\sum_{i=1}^m\left[y^{(i)}\log{a^{(i)})}+(1-y^{(i)})\log{(1-a^{(i)})}\right]\\ &=&-\frac{1}{m}\sum_{i=1}^m\left[y^{(i)}\log{h_\theta(x^{(i)})}+(1-y^{(i)})\log{(1-h_\theta(x^{(i)}))}\right]\\ &\equiv& -\frac{1}{m}\ell(\theta) \end{eqnarray*}$ where $$b\equiv\theta_0$$ and $$w\equiv(\theta_1,\ldots,\theta_d)$$. The optimization step will be carried using Gradient Descent procedures and extension which will be briefly presented in the sub-section Optimization ## 2.2 Softmax regression A softmax regression, also called a multiclass logistic regression, is used to generalized logistic regression when there are more than 2 outcome classes ($$k=1,\ldots,K$$). The outcome variable is a discrete variable $$y$$ which can take one of the $$K$$ values, $$y\in\{1,\ldots,K\}$$. The multinomial regression model is also a GLM (Generalized Linear Model) where the distribution of the outcome $$y$$ is a Multinomial$$(1,\pi)$$ where $$\pi=(\phi_1,\ldots,\phi_K)$$ is a vector with probabilities of success for each category. This Multinomial$$(1,\pi)$$ is more precisely called categorical distribution. The multinomial regression model is parameterize by $$K-1$$ parameters, $$\phi_1,\ldots,\phi_K$$, where $$\phi_i=p(y=i;\phi)$$, and $$\phi_K=p(y=K;\phi)=1-\sum_{i=1}^{K-1}\phi_i$$. By convention, we set $$\theta_K=0$$, which makes the Bernoulli parameter $$\phi_i$$ of each class $$i$$ be such that $\displaystyle\phi_i=\frac{\exp(\theta_i^Tx)}{\displaystyle\sum_{j=1}^K\exp(\theta_j^Tx)}$ where $$\theta_1,\ldots,\theta_{K-1} \ \in \Re^{d+1}$$ are the parameters of the model. This model is also called softmax regression and generalize the logistic regression. The output of the model is the estimated probability that $$p(y=i|x;\theta)$$, for every value of $$i=1,\ldots,K$$. ### 2.2.1 Multinomial regression for classification We illustrate the Multinomial model by considering three classes: red, yellow and blue. Figure 2.5: Classify for three color points clr1 <- c(rgb(1,0,0,1),rgb(1,1,0,1),rgb(0,0,1,1)) clr2 <- c(rgb(1,0,0,.2),rgb(1,1,0,.2),rgb(0,0,1,.2)) x <- c(.4,.55,.65,.9,.1,.35,.5,.15,.2,.85) y <- c(.85,.95,.8,.87,.5,.55,.5,.2,.1,.3) z <- c(1,2,2,2,1,0,0,1,0,0) df <- data.frame(x,y,z) plot(x,y,pch=19,cex=2,col=clr1[z+1]) One can use the R package to run a mutinomial regression model library(nnet) model.mult <- multinom(z~x+y,data=df) # weights: 12 (6 variable) initial value 10.986123 iter 10 value 0.794930 iter 20 value 0.065712 iter 30 value 0.064409 iter 40 value 0.061612 iter 50 value 0.058756 iter 60 value 0.056225 iter 70 value 0.055332 iter 80 value 0.052887 iter 90 value 0.050644 iter 100 value 0.048117 final value 0.048117 stopped after 100 iterations Then, the output gives a predicted probability to the three colours and we attribute the color that is the most likely. pred_mult <- function(x,y){ res <- predict(model.mult, newdata=data.frame(x=x,y=y),type="probs") apply(res,MARGIN=1,which.max) } x_grid<-seq(0,1,length=101) y_grid<-seq(0,1,length=101) z_grid <- outer(x_grid,y_grid,FUN=pred_mult) We can now visualize the three regions, the frontier being linear, and the intersection being the equiprobable case. Figure 2.6: Classifier using multinomial model image(x_grid,y_grid,z_grid,col=clr2) points(x,y,pch=19,cex=2,col=clr1[z+1]) ### 2.2.2 Likelihood of the softmax model The maximum likelihood estimation procedure consists of maximizing the log-likelihood: $\begin{eqnarray*} \ell(\theta)&=&\sum_{i=1}^m \log{p(y^{(i)|x^{(i)};\theta})}\\ &=&\sum_{i=1}^m\log{\prod_{l=1}^K\left(\frac{e^{\theta_l^Tx^{(i)}}}{\sum_{j=1}^{K}e^{\theta_j^Tx^{(i)}}}\right)^{1_{\{y^{(i)}=l\}}}} \end{eqnarray*}$ ### 2.2.3 Softmax regression as shallow Neural Network The Softmax regression model can be viewed as a shallow Neural Network. In this representation, there are $$K$$ neurons where each neuron is defined by his own set of weights $$w_i \in \Re^d$$ and a bais term $$b_i$$. The linear part is denoted by $$z_i=w_i^Tx+b_i$$ and the non linear part (activation part) is $$\sigma_i=a_i=\frac{\exp(z_i)}{\displaystyle\sum_{j=1}^K\exp(z_j)}$$. Note that the denominator of the activation part is defined using the weights from the other neurons. The output is a vector of probabilities $$(a_1,\ldots,a_K)$$ and the function is used for classification purpose: $\boxed{\hat{y}=\underset{i\in \{1,\ldots,K\}}{\textrm{argmax }\ a_i}}$ ### 2.2.4 Loss function: cross-entropy for categorical variable Let consider first one training sample $$(x,y)$$. The cross entropy loss for categorical response variable, also called Softmax Loss is defined as: $\begin{eqnarray*} CE&=&-\sum_{i=1}^K\tilde{y}_i\log p(y=i)\\ &=&-\sum_{i=1}^K\tilde{y}_i\log a_i\\ &=&-\sum_{i=1}^K\tilde{y}_i\log\left(\frac{\exp(z_i)}{\displaystyle\sum_{j=1}^K\exp(z_j)}\right) \end{eqnarray*}$ where $$\tilde{y}_i=1_{\{y=i\}}$$ is a binary variable indicating if $$y$$ is in the class $$i$$. This expression can be rewritten as $\begin{eqnarray*} CE&=&-\log \prod_{i=1}^K\left(\frac{\exp(z_i)}{\displaystyle\sum_{j=1}^K\exp(z_j)}\right)^{1_{\{y=i\}}} \end{eqnarray*}$ Then, the cost function for the $$m$$ training samples is defined as $\begin{eqnarray*} J(w,b)&=&-\frac{1}{m}\sum_{i=1}^m\log \prod_{k=1}^K\left(\frac{\exp(z^{(i)}_k)}{\displaystyle\sum_{j=1}^K\exp(z^{(i)}_j)}\right)^{1_{\{y^{(i)}=k\}}}\\ &\equiv&-\frac{1}{m}\ell(\theta) \end{eqnarray*}$ ## 2.3 Optimisation ### 2.3.1 Gradient Descent Consider unconstrained, smooth convex optimization $\underset{\theta\in \Re^d}{\text{min}}\ f(\theta),$ Algorithm : Gradient Descent 1. Choose initial point $$\theta^{(0)}\in \mathbb R^d$$ 2. Repeat $$\theta^{(k+1)}=\theta^{(k)}-\alpha.\nabla f(\theta^{(k)}), \ \ k=1,2,3,\ldots$$ 3. Stop at some point: $$||\theta^{(k+1)}-\theta^{(k)}||_2^2<\epsilon$$ Here, $$\alpha$$ is called the learning rate. Suppose that we want to find $$x$$ that minimizes: $f(x)=1.2(x-2)^2+3.2$ Figure 2.7: Closed from solution (red) f.x <- function(x) 1.2*(x-2)**2+3.2 curve(1.2*(x-2)**2+3.2,0,4,ylab="fx)") abline(v=2,col="red") In general, we cannot find a closed form solution, but can compute $$\nabla f(x)$$ simple.grad.des <- function(x0,alpha,epsilon=0.00001,max.iter=300){ tol <- 1; xold <- x0; res <- x0; iter <- 1 while (tol>epsilon & iter < max.iter){ xnew <- xold - alpha*2.4*(xold-2) tol <- abs(xnew-xold) xold <- xnew res <- c(res,xnew) iter <- iter +1 } return(res) } #result[length(result)] Convergence with a learning rate=0.01 Figure 2.8: alpha=0.01 curve(1.2*(x-2)**2+3.2,0,4,ylab="fx)") abline(v=2,col="red") points(result,f.x(result),col="blue") Convergence with a learning rate=0.83 Figure 2.9: alpha=0.83 result2 <- simple.grad.des(0,0.83,max.iter=200) #result2[length(result2)] curve(1.2*(x-2)**2+3.2,0,4,ylab="fx)") abline(v=2,col="red") points(result2,f.x(result2),col="blue",type="o") ### 2.3.2 Gradient Descent for logistic regression Given $$(x^{(i)},y^{(i)})\in \Re\times\{0,1\}$$ for $$i=1,\ldots,m$$, consider the cross-entropy loss function for this data set: $f(w)=\frac{1}{m}\sum_{i=1}^m(-y^{(i)}w^Tx^{(i)}+\log{(1+\exp(w^Tx^{(i)}))})=\sum_{i=1}^mf_i(w)$ $\nabla f(w)=\frac{1}{m}\sum_{i=1}^m(p^{(i)}(w)-y^{(i)})x^{(i)}$ where $\begin{eqnarray*} p^{(i)}(w))&=&p(Y=1|x^{(i),w})\\ &=&\exp(w^Tx^{(i)})/(1+\exp(w^Tx^{(i)})),\ \ \ i=1,\ldots,m \end{eqnarray*}$ Algorithm : Batch Gradient Descent 1. Initialize $$w=(0,\ldots,0)$$ 2. Repeat until convergence • Let $$g=(0,\ldots,0)$$ be the gradient vector • for $$i=1:m$$ do $$p^{(i)}=\exp(w^Tx^{(i)})/(1+\exp(w^Tx^{(i)}))$$ $$error^{(i)}=p^{(i)}-y_i$$ $$g=g+error^{(i)}.w^{(i)}$$ • end $$w=w-\alpha.g/m$$ 3. End repeat until convergence Note that algorithm uses all samples to compute the gradient. This approach is called batch gradient descent. ### 2.3.3 Stochastic gradient descent Algorithm : Stochastic Gradient Descent 1. Initialize $$w=(0,\ldots,0)$$ 2. Repeat until convergence Pick sample $$i$$ $$p^{(i)}=\exp(w^Tx^{(i)})/(1+\exp(w^Tx^{(i)}))$$ $$error^{(i)}=p^{(i)}-y_i$$ $$w=w-\alpha(error^{(i)}\times x^{(i)})$$ 3. End repeat until convergence Coefficient $$w$$ is updated after each sample. Remark: The gradient computation $$\nabla f(w)=\sum_{i=1}^m(p^{(i)}(w)-y^{(i)})x^{(i)}$$ is doable when $$m$$ is moderate, but not when $$m\sim 500 million$$. • One batch update costs $$O(md)$$ • One stochastic update costs $$O(d)$$ ### 2.3.4 Mini-Batches In practice, mini-batch is often used to: 1. Compute gradient based on a small subset of samples 2. Make update to coefficient vector ### 2.3.5 Example with logistic regression $$d=2$$ • Simulate some $$m$$ samples from true model: $p(Y=1|x^{(i)},w)=\frac{1}{1+\exp{(-w_1x^{(i)}_1-w_2x^{(i)})}}$ set.seed(10) m <- 5000 ;d <- 2 ;w <- c(0.5,-1.5) x <- matrix(rnorm(m*2),ncol=2,nrow=m) ptrue <- 1/(1+exp(-x%*%matrix(w,ncol=1))) y <- rbinom(m,size=1,prob = ptrue) (w.est <- coef(glm(y~x[,1]+x[,2]-1,family=binomial))) ## x[, 1] x[, 2] ## 0.557587 -1.569509 • The cross-entropy loss for this dataset Cost.fct <- function(w1,w2) { w <- c(w1,w2) cost <- sum(-y*x%*%matrix(w,ncol=1)+log(1+exp(x%*%matrix(w,ncol=1)))) return(cost) } Figure 2.10: Contour plot of the Cost function w1 <- seq(0, 1, 0.05) w2 <- seq(-2, -1, 0.05) cost <- outer(w1, w2, function(x,y) mapply(Cost.fct,x,y)) contour(x = w1, y = w2, z = cost) points(x=w.est[1],y=w.est[2],col="black",lwd=2,lty=2,pch=8) • Implementation of Batch Gradient Descent sigmoid <- function(x) 1/(1+exp(-x)) batch.GD <- function(theta,alpha,epsilon,iter.max=500){ tol <- 1 iter <-1 res.cost <- Cost.fct(theta[1],theta[2]) res.theta <- theta while (tol > epsilon & iter<iter.max) { error <- sigmoid(x%*%matrix(theta,ncol=1))-y theta.up <- theta-as.vector(alpha*matrix(error,nrow=1)%*%x) res.theta <- cbind(res.theta,theta.up) tol <- sum((theta-theta.up)**2)^0.5 theta <- theta.up cost <- Cost.fct(theta[1],theta[2]) res.cost <- c(res.cost,cost) iter <- iter +1 } result <- list(theta=theta,res.theta=res.theta,res.cost=res.cost,iter=iter,tol.theta=tol) return(result) } dim(x);length(y) ## [1] 5000 2 ## [1] 5000 Figure 2.11: Convergence Batch Gradient Descent theta0 <- c(0,-1); alpha=0.001 test <- batch.GD(theta=theta0,alpha,epsilon = 0.0000001) plot(test$res.cost,ylab="cost function",xlab="iteration",main="alpha=0.01",type="l") abline(h=Cost.fct(w.est[1],w.est[2]),col="red") Figure 2.12: Convergence of BGD contour(x = w1, y = w2, z = cost) points(x=w.est[1],y=w.est[2],col="black",lwd=2,lty=2,pch=8) record <- as.data.frame(t(test$res.theta)) points(record,col="red",type="o") • Implementation of Stochastic Gradient Descent Stochastic.GD <- function(theta,alpha,epsilon=0.0001,epoch=50){ epoch.max <- epoch tol <- 1 epoch <-1 res.cost <- Cost.fct(theta[1],theta[2]) res.cost.outer <- res.cost res.theta <- theta while (tol > epsilon & epoch<epoch.max) { for (i in 1:nrow(x)){ errori <- sigmoid(sum(x[i,]*theta))-y[i] xi <- x[i,] theta.up <- theta-alpha*errori*xi res.theta <- cbind(res.theta,theta.up) tol <- sum((theta-theta.up)**2)^0.5 theta <- theta.up cost <- Cost.fct(theta[1],theta[2]) res.cost <- c(res.cost,cost) } epoch <- epoch +1 cost.outer <- Cost.fct(theta[1],theta[2]) res.cost.outer <- c(res.cost.outer,cost.outer) } result <- list(theta=theta,res.theta=res.theta,res.cost=res.cost,epoch=epoch,tol.theta=tol) } test.SGD <- Stochastic.GD(theta=theta0,alpha,epsilon = 0.0001,epoch=10) Figure 2.13: Convergence Stochastic Gradient Descent plot(test.SGD$res.cost,ylab="cost function",xlab="iteration",main="alpha=0.01",type="l") abline(h=Cost.fct(w.est[1],w.est[2]),col="red") Figure 2.14: Convergence of Stochastic Gradient Descent contour(x = w1, y = w2, z = cost) points(x=w.est[1],y=w.est[2],col="black",lwd=2,lty=2,pch=8) record2 <- as.data.frame(t(test.SGD$res.theta)) points(record2,col="red",lwd=0.5) • Implementation of mini batch Gradient Descent Mini.Batch <- function (theta,dataTrain, alpha = 0.1, maxIter = 10, nBatch = 2, seed = NULL,intercept=NULL) { batchRate <- 1/nBatch dataTrain <- matrix(unlist(dataTrain), ncol = ncol(dataTrain), byrow = FALSE) set.seed(seed) dataTrain <- dataTrain[sample(nrow(dataTrain)), ] set.seed(NULL) res.cost <- Cost.fct(theta[1],theta[2]) res.cost.outer <- res.cost res.theta <- theta if(!is.null(intercept)) dataTrain <- cbind(1, dataTrain) temporaryTheta <- matrix(ncol = length(theta), nrow = 1) theta <- matrix(theta,ncol = length(theta), nrow = 1) for (iteration in 1:maxIter ) { if (iteration%%nBatch == 1 | nBatch == 1) { temp <- 1 x <- nrow(dataTrain) * batchRate temp2 <- x } batch <- dataTrain[temp:temp2, ] inputData <- batch[, 1:ncol(batch) - 1] outputData <- batch[, ncol(batch)] rowLength <- nrow(batch) temp <- temp + x temp2 <- temp2 + x error <- matrix(sigmoid(inputData %*% t(theta)),ncol=1) - outputData for (column in 1:length(theta)) { term <- error * inputData[, column] temporaryTheta[1, column] = theta[1, column] - (alpha * } theta <- temporaryTheta res.theta <- cbind(res.theta,as.vector(theta)) cost.outer <- Cost.fct(theta[1,1],theta[1,2]) res.cost.outer <- c(res.cost.outer,cost.outer) } result <- list(theta=theta,res.theta=res.theta,res.cost.outer=res.cost.outer) return(result) } theta0 <- c(0,-1); alpha=0.001 data.Train <- cbind(x,y) test.miniBatch <- Mini.Batch(theta=theta0,dataTrain=data.Train, alpha = 0.001, maxIter = 100, nBatch = 10, seed = NULL,intercept=NULL) ##Result frm Mini-Batch test.miniBatch$theta ## [,1] [,2] ## [1,] 0.5515469 -1.561252 Figure 2.15: Convergence Mini Batch plot(test.miniBatch$res.cost.outer,ylab="cost function",xlab="iteration",main="alpha=0.001",type="l") abline(h=Cost.fct(w.est[1],w.est[2]),col="red") Figure 2.16: Convergence of Stochastic Gradient Descent contour(x = w1, y = w2, z = cost) points(x=w.est[1],y=w.est[2],col="black",lwd=2,lty=2,pch=8) record3 <- as.data.frame(t(test.miniBatch\$res.theta)) points(record3,col="red",lwd=0.5,type="o") ## 2.4 Chain rule The univariate chain rule and the multivariate chain rule are the key concepts to calculate the derivative of cost with respect to any weight in the network. In the following a refresher of the different chain rules. ### 2.4.1 Univariate Chain rule • Univariate chain rule $\frac{\partial f(g(w))}{\partial w}=\frac{\partial f(g(w))}{\partial g(w)}.\frac{\partial g(w)}{\partial w}$ ### 2.4.2 Multivariate Chain Rule • Part I: Let $$z=f(x,y)$$, $$x=g(t)$$ and $$y=h(t)$$, where $$f,g$$ and $$h$$ are differentiable functions. Then $$z=f(x,y)=f(g(t),h(t)))$$ is a function of $$t$$, and $\begin{eqnarray*} \frac{dz}{dt} = \frac{df}{dt} &=& f_x(x,y)\frac{dx}{dt}+f_y(x,y)\frac{dy}{dt}\\ & = &\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}. \end{eqnarray*}$ • Part II: 1. Let $$z=f(x,y)$$, $$x=g(s,t)$$ and $$y=h(s,t)$$, where $$f,g$$ and $$h$$ are differentiable functions. Then $$z$$ is a function of $$s$$ and $$t$$, and $\frac{\partial z}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$ and $\frac{\partial z}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$ 1. Let $$z = f(x_1,x_2,\ldots,x_m)$$ be a differentiable function of $$m$$ variables, where each of the $$x_i$$ is a differentiable function of the variables $$t_1,t_2,\ldots,t_n$$. Then $$z$$ is a function of the $$t_i$$, and $\begin{equation*} \frac{\partial z}{\partial t_i} = \frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial t_i} + \frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial t_i} + \cdots + \frac{\partial f}{\partial x_m}\frac{\partial x_m}{\partial t_i}. \end{equation*}$ ## 2.5 Forward pass and backpropagation procedures Backpropagation is the key tool adopted by the neural network community to update the weights. This method exploits the derivative with respect to each weight $$w$$ using the chain rule (univariate and multivariate rules) ### 2.5.1 Example with the logistic model using cross-entropy loss The logistic model can be viewed as a simple neural network with no hidden layer and only a single output node with a sigmoid activation function. The sigmoid activation function $$\sigma(\cdot)$$ is applied to the linear combination of the features $$z=w^Tx+b$$ and provides the predicted value $$a$$ that represent the probability that the input $$x$$ belongs to class one. Forward pass: the forward propagation step consists to get predictions for the training samples and compute the error through a loss function in order to further adapt the weights $$w$$ and the bias $$b$$ to decrease the error. This forward pass is going through the following equations: • $$z^{(i)} = w^Tx^{(i)} + b$$ • $$a^{(i)} = \sigma(z^{(i)}) = \frac{1}{1+e^{-z^{(i)}}}\ \ \ i=1,\ldots,m$$ • $$L = J(w,b)= -\sum_{i=1}^{m} y^{(i)}\log(a^{(i)}) + (1 - y^{(i)})\log(1-a^{(i)})$$ The cost $$L=J(w,b)$$ is the error we want to reduce by adjusting the weights $$w$$ and the bais. Variations of the gradient descent algorithm are exploited to update iteratively the parameters. Thus, we have to derive the equations for the gradients on the loss function in order to propagate back the error to adapt the model parameters $$w$$ and $$b$$. Backward pass based on computation graph: The chain rule is used and generally illustrated through a computation graph: First to simplify this illustration, remind that: $\begin{eqnarray*} J(w,b) &=& \frac{1}{m} \sum_{i=1}^m \mathcal{L}(\hat{y}^{(i)}, y^{(i)})\\ &=&-\frac{1}{m}\sum_{i=1}^m\left[y^{(i)}\log{a^{(i)})}+(1-y^{(i)})\log{(1-a^{(i)})}\right]\\ \end{eqnarray*}$ Thus $\frac{\partial J(w,b)}{\partial w}=\frac{1}{m}\sum_{i=1}^m\frac{\partial \mathcal{L}(\hat{y}^{(i)}, y^{(i)})}{\partial w}$ To get $$\frac{\partial \mathcal{L}(\hat{y}^{(i)}, y^{(i)})}{\partial w}$$ the chain rule is used by considering one sample $$(x,y)$$ (the notation $$^{(i)}$$ is ommitted). Computation graphs are mainly exploited to show dependencies to derive easely the equations for the gradients. Thus to compute the gradient of the cost (lost function), one only need to go back the computation graph and multiply the gradients by each other: $\frac{\partial \mathcal{L}(\hat{y},y)}{\partial w}=\frac{\partial J(w,b)}{\partial a}.\frac{\partial a}{\partial z}.\frac{\partial z}{\partial w}$ where $$a=\sigma(z)$$ and $$z=b+w^Tx$$. • $$\frac{\partial J(w,b)}{\partial a}=-\frac{y}{a}+\frac{1-y}{1-a}=\frac{a-y}{a(1-a)}$$ • $$\frac{\partial \sigma(z)}{\partial z}= \sigma(z)(1- \sigma(z))=a(1-a)$$ • $$\frac{\partial z}{\partial w}= x$$ Thus, $\frac{\partial \mathcal{L}(\hat{y},y)}{\partial w}=x(a-y)$ and so, $\frac{\partial J(w,b)}{\partial w}=\frac{1}{m}\sum_{i=1}^mx^{(i)}(\sigma(z^{(i)})-y^{(i)})$ In the same vein, it follows $\frac{\partial J(w,b)}{\partial b}=\frac{1}{m}\sum_{i=1}^m(\sigma(z^{(i)})-y^{(i)})$ ### 2.5.2 Updating weights using Backpropagation For neural network framework, the weights are updated using gradient descent concepts $w = w - \alpha \frac{\partial J(w,b)}{\partial w}$ The main steps for updating weights are 1. Take a batch of training sample 2. Forward propagation to get the corresponding cost 3. Backpropagate the cost to get the gradients 4. update the weights using the gradients 5. Repeat step 1 to 4 for a number of iterations ## 2.6 Backpropagation for the Softmax Shallow Network ### 2.6.1 Remind some notations We consider $$K$$ class: $$y^{(i)}\in \{1,\ldots,K\}$$. Given a sample $$x$$ we want to estimate $$p(y=k|x)$$ for each $$k=1,\ldots,K$$. The softmax model is defined by $$\sigma_i=a_i=\frac{\exp(z_i)}{\displaystyle\sum_{j=1}^K\exp(z_j)}$$, with $$a_i=p(y=i|x,W)$$, $$z_i=b_i+w_i^Tx$$ and we write $$W$$ to denote all the weights of our network. Then, $$W=(w_1,\ldots,w_K)$$ a $$d\times K$$ matrix obtained by concatenating $$w_1,\ldots,w_K$$ into columns. ### 2.6.2 Chain rule using cross-entropy loss Let consider one training sample $$(x,y)$$. The cross entropy loss is $CE=-\sum_{i=1}^K\tilde{y}_i\log a_i$ where $$\tilde{y}_i=1_{\{y=i\}}$$ is a binary variable indicating if $$y$$ is in the class $$i$$. To get $$\frac{\partial CE}{\partial w_j}$$ ($$j=1,\ldots,K)$$ we need to use the multivariate chain rule: First, we derive $\begin{eqnarray*} \frac{\partial CE}{\partial z_j}&=&\sum_{k}^K\frac{\partial CE}{\partial a_k}.\frac{\partial a_k}{\partial z_j}\\ &=&\frac{\partial CE}{\partial a_j}.\frac{\partial a_j}{\partial z_j}-\sum_{k\ne j}^K\frac{\partial CE}{\partial a_k}.\frac{\partial a_k}{\partial z_j} \end{eqnarray*}$ • $$\frac{\partial CE}{\partial a_j}=-\frac{\tilde{y}_j}{a_j}$$ • if $$i=j$$ $\frac{\partial a_i}{\partial z_j}=\frac{\partial\frac{e^{z_i}}{\sum_{k=1}^K e^{z_k}}}{\partial z_j}= \frac{e^{z_i} \left(\sum_{k=1}^K e^{z_k} - e^{z_j}\right)}{\left(\sum_{k=1}^K e^{z_k}\right)^2}=\frac{ e^{z_j} }{\sum_{k=1}^K e^{z_k} } \times \frac{\left( \sum_{k=1}^K e^{z_k} - e^{z_j}\right ) }{\sum_{k=1}^K e^{z_k} }=a_i(1-a_j)$ • if $$i\ne j$$ $\frac{\partial a_i}{\partial z_j}=\frac{\partial\frac{e^{z_i}}{\sum_{k=1}^K e^{z_k}}}{\partial z_j}=\frac{0 - e^{z_j}e^{z_i}}{\left( \sum_{k=1}^K e^{z_k}\right)^2}=\frac{- e^{z_j} }{\sum_{k=1}^K e^{z_k} } \times \frac{e^{z_i} }{\sum_{k=1}^K e^{z_k} }=- a_j.a_i$ So we can rewrite it as $\frac{\partial a_i}{\partial z_j} = \left\{ \begin{array}{ll} a_i(1-a_j) & if & i=j \\ -a_j.a_i & if & i\neq j \end{array}\right.$ Thus, we get $\begin{eqnarray*} \frac{\partial CE}{\partial z_j}&=&\frac{\partial CE}{\partial a_j}.\frac{\partial a_j}{\partial z_j}-\sum_{k\ne j}^K\frac{\partial CE}{\partial a_k}.\frac{\partial a_k}{\partial z_j}\\ &=&-\tilde{y}_j(1-a_j)+\sum_{k\ne j}^K\tilde{y}_ja_k\\ &=&-\tilde{y}_j+a_j\sum_{k}^K\tilde{y}_k=a_j-\tilde{y}_j\\ \end{eqnarray*}$ We can now derive the gradient for the weights as: $\begin{eqnarray*} \frac{\partial CE}{\partial w_j}&=&\sum_{k}^K\frac{\partial CE}{\partial z_k}.\frac{\partial z_k}{\partial w_j}\\ &=&(a_j-\tilde{y}_j)x \end{eqnarray*}$ In the same way, we get $\begin{eqnarray*} \frac{\partial CE}{\partial b_j}&=&\sum_{k}^K\frac{\partial CE}{\partial z_k}.\frac{\partial z_k}{\partial b_j}\\ &=&(a_j-\tilde{y}_j) \end{eqnarray*}$ ### 2.6.3 Computation Graph could help Let consider a simple example with $$K=3$$ ($$y\in\{1,2,3\}$$) and two features ($$x_1$$ and $$x_2$$). The computational graph for this softmax neural network model help us to visualize dependencies between nodes and then to derive the gradient of the cost (loss) in respect to each parameter ($$w_j\in\ \Re^{2}$$ and $$b_j \in \Re$$, $$j=1,2,3$$). Let’s write as an example for $$\frac{\partial L}{\partial w_{2,1}}$$: \begin{align*} \frac{\partial L}{\partial w_{2,1}} & = \sum_{i=1}^3 \left (\frac{\partial L}{\partial a_i} \right) \left (\frac{\partial a_i}{\partial z_2} \right) \left(\frac{\partial z_2}{\partial w_{2,1}} \right ) \\ &= \left (\frac{\partial L }{\partial a_1} \right) \left (\frac{\partial a_1}{\partial z_2} \right) \left(\frac{\partial z_2}{\partial w_{2,1}} \right ) + \left (\frac{\partial L }{\partial a_2} \right) \left (\frac{\partial a_2}{\partial z_2} \right) \left(\frac{\partial z_2}{\partial w_{2,1}} \right ) + \left (\frac{\partial L}{\partial a_3} \right) \left (\frac{\partial a_3}{\partial z_2} \right) \left(\frac{\partial z_2}{\partial w_{2,1}} \right ) \end{align*} In fact, we are summing up the contribution of the change of $$w_{2,1}$$ over different “paths” (in red from figure above). When we change $$w_{2,1}$$; $$\ a_1\ a_2$$ and $$a_3$$ changes as a result. Then, the change of $$\ a_1\ a_2$$ and $$a_3$$ affects $$L$$. We sum up all the changes $$w_{2,1}$$ produced over $$\ a_1\ a_2$$ and $$a_3$$ to $$L$$. import numpy as np import matplotlib.pyplot as plt import sympy from scipy import optimize Page built: 2021-03-04 using R version 4.0.3 (2020-10-10)

https://socratic.org/questions/what-is-the-sum-of-all-odd-numbers-between-0-and-100

# What is the sum of all odd numbers between 0 and 100? ##### 1 Answer Sep 17, 2015 First, notice an interesting pattern here: $1 , 4 , 9 , 16 , 25 , \ldots$ The differences between perfect squares (starting at $1 - 0 = 1$) is: $1 , 3 , 5 , 7 , 9 , \ldots$ The sum of $1 + 3 + 5 + 7 + 9$ is $25$, the ${5}^{\text{th}}$ nonzero square. Let's take another example. You can quickly prove that: $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100$ There are $\frac{19 + 1}{2} = 10$ odd numbers here, and the sum is ${10}^{2}$. Therefore, the sum of $1 + 3 + 5 + \ldots + 99$ is simply: ${\left(\frac{99 + 1}{2}\right)}^{2} = \textcolor{b l u e}{2500}$ Formally, you can write this as: $\textcolor{g r e e n}{{\sum}_{n = 1}^{N} \left(2 n - 1\right) = 1 + 3 + 5 + \ldots + \left(2 N - 1\right) = {\left(\frac{N + 1}{2}\right)}^{2}}$ where $N$ is the last number in the sequence and $n$ is the index of each number in the sequence. So, the ${50}^{\text{th}}$ number in the sequence is $2 \cdot 50 - 1 = 99$, and the sum all the way up to that is ${\left(\frac{99 + 1}{2}\right)}^{2} = 2500$.

https://www.xarg.org/proof/svg-arc-to-gcode-g2-and-g3/

# Proof: SVG Arc to GCode G2 and G3 The SVG arc command consists of A rx ry x-axis-rotation large-arc-flag sweep-flag x y • rx: Ellipse radius of x-axis • ry: Ellipse radius of y-axis • x-axis-rotation: Coordinate system rotation in degrees • large-arc-flag: Flag if large or small circle should be taken • sweep-flag: Flag on which side of the line between start and end should the circle is drawn • x: The final x-coordinate • y: The final y-coordinate For the case that $$r_x\neq r_y$$, we can only sample from the path and interpolate it linearly. But for the case that $$r_x=r_y$$, it’s possible to improve the gcode by using G2 and G3. In general, the syntax of G2 and G3 are G2 I<offset> J<offset> R<radius> [X<pos>] [Y<pos>] G3 I<offset> J<offset> R<radius> [X<pos>] [Y<pos>] • I: An offset from the current X position to use as the arc center • Y: An offset from the current Y position to use as the arc center • R: The radius from the current XY position to use as the arc center • X: The final x-coordinate • Y: The final y-coordinate The combination (I, J) or R are exclusive. As an example, an arc can be drawn like this with (I, J) offset or via radius R: ## Derivation In this derivation, the use of the R parameter is ignored, and the gcode is fed with the (I, J) tuple. Let’s say $$\mathbf{S}$$ is the absolute starting point of the SVG arc path and $$\mathbf{E}$$ the absolute ending point of the SVG arc path. The points are connected with the vector $$\mathbf{a}=\mathbf{E}-\mathbf{S}$$. The point $$\mathbf{M} = \frac{1}{2}(\mathbf{S}+\mathbf{E})$$ is the mid-point between start and end. On this point, we can construct an orthogonal vector $$\mathbf{v}$$ to our desired center $$\mathbf{C}$$. The length $$|\mathbf{v}|$$ can be determined using Pythagorean theorem: $|\mathbf{v}| = \sqrt{r^2 - \frac{1}{4}|\mathbf{a}|^2}$ Now the vector $$\mathbf{v}$$ can be found with the perp operator: $\mathbf{v} = \mathbf{\hat{a}}^\perp |\mathbf{v}|$ Finally, the center (I, J) can be found with $\begin{array}{rl} (I, J) =& \mathbf{M} \pm \mathbf{v} - \mathbf{S}\\ =& \mathbf{M} \pm \mathbf{\hat{a}}^\perp |\mathbf{v}| - \mathbf{S}\\ =& \frac{1}{2}(\mathbf{S}+\mathbf{E}) - \mathbf{S} \pm \frac{\mathbf{{a}}^\perp}{|\mathbf{a}|} \sqrt{r^2 - \frac{1}{4}|\mathbf{a}|^2}\\ =& \frac{1}{2}(\mathbf{E} - \mathbf{S}) \pm \mathbf{{a}}^\perp \sqrt{\frac{r^2}{|\mathbf{a}|^2} - \frac{1}{4}}\\ =& \frac{1}{2}\left(\mathbf{a} \pm \mathbf{{a}}^\perp \sqrt{\frac{4r^2}{\mathbf{a}\cdot\mathbf{a}} - 1}\right)\\ \end{array}$ For which the positive part of the equation is used when the sweep-flag and large-arc-flag differ $$f_S\neq f_A$$ and the negative part is used for $$f_S=f_A$$. The remaining question is to decide if we need to draw the circle clockwise or counter-clockwise, which is determined by the direction the circle is drawn - or simply if the sweep-flag $$f_S=1$$. ## Pseudo-Code func arcToGCode(S, E, r, fS, fA) { a = { x: E.x - S.x, y: E.y - S.y } aP = fS != fA ? { x: -a.y, y: a.x } : { x: a.y, y: -a.x } w = sqrt(4 * r * r / (a.x * a.x + a.y * a.y) - 1) I = (a.x + aP.x * w) / 2 J = (a.y + aP.y * w) / 2 if (fS != 1) return "G2 X" + E.x + " Y" + E.y + " I" + I + " J" + J else return "G3 X" + E.x + " Y" + E.y + " I" + I + " J" + J }

https://math.stackexchange.com/questions/1967156/question-about-induction

Prove by induction $w_k = w_{k−2} + k$, for all integers $k \ge 3, w_1 = 1,w_2 = 2$ has an explicit formula $$w_n = \begin{cases} \frac{(n+1)^2}{4}, & \text{if n is odd} \\ \frac n2(\frac n2 + 1), & \text{if n is even} \end{cases}$$ Inductive step for when $n$ is odd: Suppose $w_k = \frac{(k+1)^2}{4}$ if $k$ is odd. Then by definition of $w$, we have $w_{k + 2} = w_k + k + 2 = \frac{(k+1)^2}{4} + k + 2 = \frac {k^2 + 2k + 1}{4} + k + 2= \frac {k^2 + 6k + 8}{4} = \frac {(k +3)^2}{4}$ if $k + 2$ is odd. Is it important that we prove $w_{k + 1} = \frac{(k+2)^2}{4}$ if $k + 1$ is odd or is the proof for $w_{k + 2} = \frac{(k+3)^2}{4}$ if $k + 2$ is enough? The beginning is clear. After that, for the (strong) induction step, you want to show that $w_n$ is given by the formula specified, provided this is the case for $w_1,w_2,\ldots, w_{n-1}$. By the nature of the recursion, we need only that the formula holds for $w_{n-2}$ (which is among the given $w_i$ because we also assume $n>2$ for the induction step!). AS $n-2$ has the same parity as $n$, we conclude that $$w_n=w_{n-2}+n=\begin{cases}\frac{(n-2+1)^2}{4}+n,&\text{if n and n-2 are odd}\\\frac{n-2}2(\frac{n-2}2+1)+n,&\text{if n and n-2 are odd}\end{cases}$$ Simple transformations should bring the desired result ...

https://yourgametips.com/miscellaneous/what-is-the-probability-of-drawing-an-ace-from-a-deck-of-52-cards/

# What is the probability of drawing an ace from a deck of 52 cards? ## What is the probability of drawing an ace from a deck of 52 cards? The probability of picking up an ace in a 52 deck of cards is 4/52 since there are 4 aces in the deck. ## What is the probability of getting either a spade or a jack when drawing a single card from a deck of 52 cards? There are 4 jacks in the deck and 13 spades. However 1 jack is a spade so we have a total of 16 cards which are either a jack or a spade. Therefore there are 52−16=36 cards which are not a jack or a spade. Thus the probability is 36/52. What is the probability of drawing a red card or ace? The probability of drawing a red card is 1/2 as half the cards are red. So we need the sum 1/2 + 4/52 but we have counted red aces in both of these so we need to subtract the probability of drawing a red ace. So p(draw an ace or a red card) = 4/52 + 26/52 – 2/52 = 28/52. What is the probability of drawing two aces from a deck of 52 cards? Thus, the chance of drawing an ace on each of two draws is 4/52 × 3/51, or 1/221.

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On ideals of rings of fractions and rings of polynomials nai, yuan ting and zhao, dongsheng, kodai mathematical journal, 2015. A ring all of whose ideals are principal is called a principal ideal ring, two important cases are z and kx, the polynomial ring over a field k. In s, we have studied those prime left principal ideal rings, especially domains, which contain an isomorphic copy of their left quotient rings and we have shown. Every commutative unital algebraically closed or principal ideal ring is associate. Equivalently, it is a right principal ideal or a twosided principal ideal of. Similarity classes of 3x3 matrices over a local principal ideal ring. In this paper similarity classes of three by three matrices over a local principal ideal commutative ring are analyzed. In fact, we prove that rx is a principal ideal ring if and only if r is a finite direct product of finite fields. Examples of principal ideal rings include the ring of integers, the ring of polynomials over a field, the ring of skew polynomials over a field with an automorphism the elements of have the form, the addition of these. The term also has another, similar meaning in order theory, where it refers to an order ideal in a poset generated by a single element. The right and left ideals of this form, generated by one element, are called principal ideals. An ideal icris a principal ideal if i haifor some a2r. Minimal monomial reductions and the reduced fiber ring of an extremal ideal singla, pooja, illinois journal of. In mathematics, a principal right left ideal ring is a ring r in which every right left ideal is of the form xr rx for some element x of r. In mathematics, specifically ring theory, a principal ideal is an ideal in a ring that is generated by a single element of through multiplication by every element of. The imbedding of a ring as an ideal in another ring johnson, r. It is well known that every euclidean ring is a principal ideal ring. A nonzero ring in which 0 is the only zero divisor is called an integral domain. A subring a of a ring r is called a twosided ideal of r if for every r 2 r and every a 2 a, ra 2 a and ar 2 a. Principal ideal domains appear in the following chain of class inclusions. Synonyms smallest ideal that contains a given element. An ideal a of r is a proper ideal if a is a proper subset of r. The mathematical system which seems most satisfactory as an abstraction of the system of rational integers is the principal ideal ring. Some examples of principal ideal domain which are not euclidean and some other counterexamples veselin peric1, mirjana vukovic2 abstract. Principal ideal ring, polynomial ring, finite rings. Finite commutative rings are interesting objects of ring theory and have many. Commutative ring theorydivisibility and principal ideals. Counterexamples exist under the rings r of integral algebraic. If r is an integral domain then the polynomial ring rx is also. Associative rings and algebras in which all right and left ideals are principal, i. A principal ideal ring that is not a euclidean ring. It is also known for a very long time that the converse is not valid. Since every principal ideal domain commutative or not is a fir, we find in parti cular that firs include a free products of fields over a given field, b free. Every commutative ring embeds into an associate ring. Consider a principal ideal ring r and the ring homomorphism r s. Any ideal that is not contained in any proper ideal i. Let r be the ring zn of integers modulo n, where n may be prime or composite. More generally, a principal ideal ring is a nonzero commutative ring whose. When this is satisfied for both left and right ideals, such as the case when r is a commutative ring, r can be called a principal ideal ring, or simply. A ring ris a principal ideal domain pid if it is an integral domain 25. Left principal ideal domains a ring r is a left principal ideal. The key point will be that the principal ideals corresponds to the element and its associates, and the non principal ideals will correspond to ideal elements of r. Let r \displaystyle r be a commutative ring, and let a, b. Show that the homomorphic image of a principal ideal ring. Any ring has two ideals, namely the zero ideal 0 and r, the whole ring. In mathematics, a principal ideal domain, or pid, is an integral domain in which every ideal is principal, i. Proposition characterisation of divisibility by principal ideals. An integral domain r is said to be a euclidean ring iffor every x. 274 896 953 1227 1316 1457 1433 800 777 1158 55 884 1107 790 318 1202 1443 1596 388 1083 1275 208 18 846 1444 1526 96 376 1044 368 1151 1120 1377 897 214 941 518 277 1010 520 1488 584 1044 260 461 288 1092 104